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How to Solve a Quadratic Equation: A Step-by-Step Guide
Last Updated: May 27, 2024 Fact Checked
Factoring the Equation
Using the quadratic formula, completing the square, practice problems and answers, expert q&a.
This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,431,733 times.
A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.
Quadradic Formula for Solving Equations
- Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. [3] X Research source
- 3x = -1 ..... by subtracting
- 3x/3 = -1/3 ..... by dividing
- x = -1/3 ..... simplified
- x = 4 ..... by subtracting
- x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.
- So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.
- 4x 2 - 5x - 13 = x 2 -5
- 4x 2 - x 2 - 5x - 13 +5 = 0
- 3x 2 - 5x - 8 = 0
- {-b +/-√ (b 2 - 4ac)}/2
- {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
- {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)
- {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
- {5 +/-√(25 + 96)}/6
- {5 +/-√(121)}/6
- (5 + 11)/6 = 16/6
- (5-11)/6 = -6/6
- x = (-1, 8/3)
- 2x 2 - 9 = 12x =
- In this equation, the a term is 2, the b term is -12, and the c term is -9.
- 2x 2 - 12x - 9 = 0
- 2x 2 - 12x = 9
- 2x 2 /2 - 12x/2 = 9/2 =
- x 2 - 6x = 9/2
- -6/2 = -3 =
- (-3) 2 = 9 =
- x 2 - 6x + 9 = 9/2 + 9
- x = 3 + 3(√6)/2
- x = 3 - 3(√6)/2)
- If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [14] X Research source Thanks Helpful 0 Not Helpful 0
- If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 3 Not Helpful 0
- As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0
You Might Also Like
- ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
- ↑ https://www.mathsisfun.com/algebra/factoring-quadratics.html
- ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
- ↑ https://www.cuemath.com/algebra/quadratic-equations/
- ↑ https://www.purplemath.com/modules/solvquad4.htm
- ↑ http://www.purplemath.com/modules/quadform.htm
- ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
- ↑ http://www.mathsisfun.com/algebra/completing-square.html
- ↑ https://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm
About This Article
To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No
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Solving Quadratics by the Quadratic Formula
How to solve quadratic equations using the quadratic formula.
There are times when we are stuck solving a quadratic equation of the form [latex]a{x^2} + bx + c = 0[/latex] because the trinomial on the left side can’t be factored out easily. It doesn’t mean that the quadratic equation has no solution. At this point, we need to call upon the straightforward approach of the quadratic formula to find the solutions of the quadratic equation or put simply, determine the values of [latex]x[/latex] that can satisfy the equation.
In order use the quadratic formula, the quadratic equation that we are solving must be converted into the “standard form”, otherwise, all subsequent steps will not work. The goal is to transform the quadratic equation such that the quadratic expression is isolated on one side of the equation while the opposite side only contains the number zero, [latex]0[/latex].
Take a look at the diagram below.
In this convenient format, the numerical values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easily identified! Upon knowing those values, we can now substitute them into the quadratic formula then solve for the values of [latex]x[/latex].
- The Quadratic Formula
- Where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are the coefficients of an arbitrary quadratic equation in the standard form, [latex]a{x^2} + bx + c = 0[/latex].
Slow down if you need to. Be careful with every step while simplifying the expressions. This is where common mistakes usually happen because students tend to “relax” which results to errors that could have been prevented, such as in the addition, subtraction, multiplication and/or division of real numbers.
Examples of How to Solve Quadratic Equations by the Quadratic Formula
Example 1 : Solve the quadratic equation below using the Quadratic Formula.
By inspection, it’s obvious that the quadratic equation is in the standard form since the right side is just zero while the rest of the terms stay on the left side. In other words, we have something like this
This is great! What we need to do is simply identify the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] then substitute into the quadratic formula.
That’s it! Make it a habit to always check the solved values of [latex]x[/latex] back into the original equation to verify.
Example 2 : Solve the quadratic equation below using the Quadratic Formula.
This quadratic equation is absolutely not in the form that we want because the right side is NOT zero. I need to eliminate that [latex]7[/latex] on the right side by subtracting both sides by [latex]7[/latex]. That takes care of our problem. After doing so, solve for [latex]x[/latex] as usual.
The final answers are [latex]{x_1} = 1[/latex] and [latex]{x_2} = – {2 \over 3}[/latex].
Example 3 : Solve the quadratic equation below using the Quadratic Formula.
This quadratic equation looks like a “mess”. I have variable [latex]x[/latex]’s and constants on both sides of the equation. If we are faced with something like this, always stick to what we know. Yes, it’s all about the Standard Form. We have to force the right side to be equal to zero. We can do just that in two steps.
I will first subtract both sides by [latex]5x[/latex], and followed by the addition of [latex]8[/latex].
Values we need:
[latex]a = – 1[/latex], [latex]b = – \,8[/latex], and [latex]c = 2[/latex]
Example 4 : Solve the quadratic equation below using the Quadratic Formula.
Well, if you think that Example [latex]3[/latex] is a “mess” then this must be even “messier”. However, you’ll soon realize that they are really very similar.
We first need to perform some cleanup by converting this quadratic equation into standard form. Sounds familiar? Trust me, this problem is not as bad as it looks, as long as we know what to do.
Just to remind you, we want something like this
Therefore, we must do whatever it takes to make the right side of the equation equal to zero. Since we have three terms on the right side, it follows that three steps are required to make it zero.
The solution below starts by adding both sides by [latex]3{x^2}[/latex], followed by subtraction of [latex]3x[/latex], and finally the addition of [latex]5[/latex]. Done!
After making the right side equal to zero, the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easy to identify. Plug those values into the quadratic formula, and simplify to get the final answers!
Example 5 : Solve the quadratic equation below using the Quadratic Formula.
First, we need to rewrite the given quadratic equation in Standard Form, [latex]a{x^2} + bx + c = 0[/latex].
- Eliminate the [latex]{x^2}[/latex] term on the right side.
- Eliminate the [latex]x[/latex] term on the right side.
- Eliminate the constant on the right side.
After getting the correct standard form in the previous step, it’s now time to plug the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] into the quadratic formula to solve for [latex]x[/latex].
- From the converted standard form, extract the required values.
[latex]a = 1[/latex], [latex]b = – \,4[/latex], and [latex]c = – \,14[/latex]
- Then evaluate these values into the quadratic formula.
You might also like these tutorials:
- Quadratic Formula Practice Problems with Answers
- Solving Quadratic Equations by Square Root Method
- Solving Quadratic Equations by Factoring Method
- Solving Quadratic Equations by Completing the Square
Quadratic Expressions
Quadratic expression is an expression with the variable with the highest power of 2. The word quadratic is derived from the word quad which means square. The expression should have the power of two and not higher or lower. Graphically a quadratic expression describes the path followed by a parabola, and it helps in finding the height and time of flight of a rocket.
In this mini-lesson, we will explore the world of quadratic expressions. You will get to learn how to solve quadratic expressions using the discriminant, simplifying quadratic expressions, factorizing expressions, quadratic expressions graphs, and other interesting facts around the topic.
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Definition of Quadratic Expressions
An expression of the form ax 2 + bx + c, where a ≠ 0 is called a quadratic expression. In other words, any expression with the variables of the highest exponent or the expression's degree as 2 is a quadratic expression The standard form of a quadratic expression looks like this
Here are some examples of expressions:
| Expressions | Values of a, b, and c | |
|---|---|---|
| 1. | 8x + 7x - 1 | a = 8 , b = 7 , c = -1 |
| 2. | 4x - 9 | a = 4 , b = 0 , c = -9 |
| 3. | x + x | a = 1 , b = 1 , c = 0 |
| 4. | 6x - 8 | a = 0 , b = 6 , c = -8 |
Look at example 4, here a = 0. Therefore, it is not a quadratic expression. In fact, this is a linear expression. The remaining expressions are quadratic expression examples. The standard form of the standard expression in variable x is ax 2 + bx + c. But always remember that 'a' is a non-zero value and sometimes a quadratic expression is not written in its standard form.
Properties of Quadratic Expression
Listed below are a few important properties to keep in mind while identifying quadratic expressions.
- The expression is usually written in terms of x, y, z, or w. In the alphabets, the letters, in the end, are written for variables whereas the letters, in the beginning, are used for numbers.
- The variable 'a' in a quadratic expression raised to the power of 2 cannot be zero. If a = 0 then x 2 will be multiplied by zero and therefore, it would not be a quadratic expression anymore. Variable b or c in the standard form can be 0 but 'a' cannot.
- In a quadratic expression, it is not necessary that all the terms are positive. The standard form is written in a positive form. However, if the numbers are negative the term will also be negative.
- The terms in a quadratic expression are usually written with the power of 2 first, the power of 1 next, and the number in the end.
Ways of Writing a Quadratic Expression
Writing any expression or equation into a quadratic standard form, we need to follow these three methods.
- Move an expression by -1 if the equation starts with a negative value.
- Expand the terms and simply.
- Multiply the factors.
Mentioned below are a few examples:
| (x-1)(x+2) | Multiply the factors (x-1) and (x+2) | x + x - 2 |
| -x + 3x - 1 | Move expression by -1 | x - 3x + 1 |
| 5x(x+3) - 12x | Expand the term 5x(x+3) | 5x + 3x |
| x - x(x + x - 3) | Expand the term x(x + x - 3) and simplify | x - 3x |
Graphing a Quadratic Expressions
The graph of the quadratic expression ax 2 + bx + c = 0 can be obtained by representing the quadratic expression as a function y = ax 2 + bx + c. Further on solving and substituting values for x, we can obtain values of y, we can obtain numerous points. These points can be presented in the coordinate axis to obtain a parabola-shaped graph for the quadratic expression.
The point where the graph cuts the horizontal x-axis is the solution of the quadratic expression. These points can also be algebraically obtained by equalizing the y value to 0 in the function y = ax 2 + bx + c and solving for x. This is how the quadratic expression is represented on a graph. This curve is called a parabola .
Factorizing Quadratic Expressions
The factorizing of a quadratic expression is helpful for splitting it into two simpler expressions, which on multiplying, gives back the original quadratic expression. The aim of factorization is the break down the expression of higher degrees into expressions of lower degrees. Here the quadratic expression is of the second degree and is split into two simple linear expressions. The process of factorization of a quadratic equation includes the first step of splitting the x term, such that the product of the coefficients of the x-term is equal to the constant term. Further, the new expression after splitting the x term has four terms in the quadratic expression. Here some of the terms are taken commonly to obtain the final factors of the quadratic expression. Let us understand the process of factorizing a quadratic expression through an example.
Example 1: 4x - 12x 2
Given any quadratic expression, first, check for common factors, i.e. 4x and 12x 2
We can observe that 4x is a common factor. Let’s take that common factor from the quadratic expression.
4x - 12x 2 = 4x(1 - 3x)
Thus, by simplifying the quadratic expression, we get the expression, 4x - 12x 2 .
Quadratic Expressions Formula
Solving a quadratic expression is possible if we are able to convert it into a quadratic equation by equalizing it to zero. The values of the variable x which satisfy the quadratic expression and equalize it to zero are called the zeros of the equation. Some of the expressions cannot be easily solved by the method of factorizing. The quadratic formula is here to help. The quadratic formula is also known as "Quadranator." Quadranator alone is enough to solve all quadratic expression problems. The quadratic expressions formula is as follows.
Discriminant
The discriminant is an important part of the quadratic expression formula. The value of the discriminant is (b 2 - 4ac). This is called a discriminant because it discriminates the zeroes of the quadratic expression based on its sign. Many an instance we wish to know more about the zeros of the equation, before calculating the roots of the expression. Here the discriminant value is useful. Based on the values of the discriminant the nature of the zeros of the equation or the roots of the equation can be predicted.
The nature of the roots of the quadratic expression based on the discriminant value is as follows.
- When the quadratic expression has two real and distinct roots: b 2 - 4ac > 0
- When the quadratic expression has equal roots: b 2 - 4ac = 0
- When the quadratic expression does not have any roots or has imaginary roots: b 2 - 4ac < 0
Related Topics
The following topics are related to quadratic expressions.
- Quadratic Equation
- Graphing a Quadratic Function
- Polynomial Functions
- Algebraic Expressions
- Domain and Range of a Parabola
Examples on Quadratic Expressions
Example 1: Jack shows a quadratic expression. The quadratic expression is x 2 + 3x - 4. Can you find the value of expression at x = -2?
Suppose we have an expression p(x) in variable (x).
The value of expression at (k) is given by (p(k)).
To find the value of expression at (x = -2), substitute -2 for (x) in the given expression.
(-2) 2 + 3(-2) -4 = 4- 6 -4 = -6
Therefore the value of expression at x = 2 is -6.
Example 2: Flora loves collecting flowers. She wants to choose a number such that it is the solution to the quadratic expression \(x^{2}-8x+12\). Can you help her choose the correct number?
Compare the quadratic expression \(x^{2}-8x+12\) with the standard form of quadratic expression \(ax^{2}+bx+c\).
We get, \(a=1\), \(b=-8\) and the constant \(c=12\)
Now calculate the solution of the quadratic expression by using the quadratic formula.
\[\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(12)}}{2}\\&=\frac{8\pm\sqrt{64-48}}{2}\\&=\frac{8\pm\sqrt{16}}{2}\\&=\frac{8\pm4}{2}\\x&=6,2\end{aligned}\]
The number 2 is not on the flowers. Therefore, she will choose the flower on which number 6 is written.
Therefore the correct number is 6.
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Practice Questions on Quadratic Expressions
Faqs on quadratic expressions, what is a quadratic expression in math.
An expression of the form ax 2 + bx + c, where a ≠ 0 is called a quadratic expression in variable x.
What Are the 3 Forms of a Quadratic Expression?
The three forms of quadratic expressions are
- Standard form
- Factored form
- Vertex form
What Does it Mean to Factor a Quadratic Expression?
The factored form of a quadratic expression is to split the quadratic expression into a product of two factors. A quadratic expression is the multiplication of two linear expressions, and the process of factorization is to find the two factors of the quadratic expression..
Which Is a Quadratic Function?
A function of the form f(x) = ax 2 + bx + c, where a ≠ 0 is called a quadratic function in variable x. Similar to a function , a quadratic expression also has a domain and a range value.
What Does Quadratic Mean in a Quadratic Expression?
The word "quadratic" is derived from the word "quad" which means square. In the quadratic expression ax 2 + bx + c, we have the term x 2 which refers to the term quadratic.
How Do you Know If a Polynomial is a Quadratic Expression?
The polynomial expression having a maximum of x 2 term is termed as a quadratic expression. The general form of a quadratic expression is ax 2 + bx + c, and If the coefficient of x 2 is non-zero, then the expression is a quadratic expression.
What Are the Characteristics of a Quadratic Expression?
The characteristics of a quadratic expression are:
- The standard form of the quadratic expression is ax 2 + bx + c, where a ≠ 0.
- A quadratic expression has at most two zeroes.
- The curve of the quadratic expression is in the form of a parabola.
What Are the Terms in a Quadratic Expression?
A quadratic expression generally has three terms. The x 2 term, the x term, and the constant term. The standard form of a quadratic expression is of the form ax 2 + bx + c, and a is a non-zero value. Here a, b are called the coefficients and c is the constant term.
How do you write Quadratic Expressions in Standard Form?
The standard form of a quadratic equation is ax 2 + bx + c, where a ≠ 0 in variable x.
How Do you Simplify a Quadratic Expression?
Quadratic equations can be simplified by the process of factorization. Further, the other methods of solving a quadratic equation are by using the formula, and by the method of finding squares. For a quadratic expression of the form x 2 + (a + b)x + ab, the process of factorization gives the following simplified factors (x + a)(x + b).
- Parallelogram
- Quadrilateral
- Parallelepiped
- Tetrahedron
- Dodecahedron
- Fraction Calculator
- Mixed Fraction Calculator
- Greatest Common Factor Calulator
- Decimal to Fraction Calculator
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- Place Value
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Last modified on August 3rd, 2023
Quadratic Equation Word Problems
Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them:
- By Completing the Square
- By Factoring
- By Quadratic Formula
- By graphing
For each process, follow the following typical steps:
- Make the equation
- Solve for the unknown variable using the appropriate method
- Interpret the result
The product of two consecutive integers is 462. Find the numbers?
Let the numbers be x and x + 1 According to the problem, x(x + 1) = 483 => x 2 + x – 483 = 0 => x 2 + 22x – 21x – 483 = 0 => x(x + 22) – 21(x + 22) = 0 => (x + 22)(x – 21) = 0 => x + 22 = 0 or x – 21 = 0 => x = {-22, 21} Thus, the two consecutive numbers are 21 and 22.
The product of two consecutive positive odd integers is 1 less than four times their sum. What are the two positive integers.
Let the numbers be n and n + 2 According to the problem, => n(n + 2) = 4[n + (n + 2)] – 1 => n 2 + 2n = 4[2n + 2] – 1 => n 2 + 2n = 8n + 7 => n 2 – 6n – 7 = 0 => n 2 -7n + n – 7 = 0 => n(n – 7) + 1(n – 7) = 0 => (n – 7) (n – 1) = 0 => n – 7 = 0 or n – 1 = 0 => n = {7, 1} If n = 7, then n + 2 = 9 If n = 1, then n + 1 = 2 Since 1 and 2 are not possible. The two numbers are 7 and 9
A projectile is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet tower. If height after t seconds is reprented by h(t) = -16t 2 + 64t + 96. Find the maximum height the projectile reaches. Also, find the time it takes to reach the highest point.
Since the graph of the given function is a parabola, it opens downward because the leading coefficient is negative. Thus, to get the maximum height, we have to find the vertex of this parabola. Given the function is in the standard form h(t) = a 2 x + bx + c, the formula to calculate the vertex is: Vertex (h, k) = ${\left\{ \left( \dfrac{-b}{2a}\right) ,h\left( -\dfrac{b}{2a}\right) \right\}}$ => ${\dfrac{-b}{2a}=\dfrac{-64}{2\times \left( -16\right) }}$ = 2 seconds Thus, the time the projectile takes to reach the highest point is 2 seconds ${h\left( \dfrac{-b}{2a}\right)}$ = h(2) = -16(2) 2 – 64(2) + 80 = 144 feet Thus, the maximum height the projectile reaches is 144 feet
The difference between the squares of two consecutive even integers is 68. Find the numbers.
Let the numbers be x and x + 2 According to the problem, (x + 2) 2 – x 2 = 68 => x 2 + 4x + 4 – x 2 = 68 => 4x + 4 = 68 => 4x = 68 – 4 => 4x = 64 => x = 16 Thus the two numbers are 16 and 18
The length of a rectangle is 5 units more than twice the number. The width is 4 unit less than the same number. Given the area of the rectangle is 15 sq. units, find the length and breadth of the rectangle.
Let the number be x Thus, Length = 2x + 5 Breadth = x – 4 According to the problem, (2x + 5)(x – 4) = 15 => 2x 2 – 8x + 5x – 20 – 15 = 0 => 2x 2 – 3x – 35 = 0 => 2x 2 – 10x + 7x – 35 = 0 => 2x(x – 5) + 7(x – 5) = 0 => (x – 5)(2x + 7) = 0 => x – 5 = 0 or 2x + 7 = 0 => x = {5, -7/2} Since we cannot have a negative measurement in mensuration, the number is 5 inches. Now, Length = 2x + 5 = 2(5) + 5 = 15 inches Breadth = x – 4 = 15 – 4 = 11 inches
A rectangular garden is 50 cm long and 34 cm wide, surrounded by a uniform boundary. Find the width of the boundary if the total area is 540 cm².
Given, Length of the garden = 50 cm Width of the garden = 34 cm Let the uniform width of the boundary be = x cm According to the problem, (50 + 2x)(34 + 2x) – 50 × 34 = 540 => 4x 2 + 168x – 540 = 0 => x 2 + 42x – 135 = 0 Since, this quadratic equation is in the standard form ax 2 + bx + c, we will use the quadratic formula, here a = 1, b = 42, c = -135 x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$ => ${\dfrac{-42\pm \sqrt{\left( 42\right) ^{2}-4\times 1\times \left( -135\right) }}{2\times 1}}$ => ${\dfrac{-42\pm \sqrt{1764+540}}{2}}$ => ${\dfrac{-42\pm \sqrt{2304}}{2}}$ => ${\dfrac{-42\pm 48}{2}}$ => ${\dfrac{-42+48}{2}}$ and ${\dfrac{-42-48}{2}}$ => x = {-45, 3} Since we cannot have a negative measurement in mensuration the width of the boundary is 3 cm
The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.
Let the length of the other two sides be x and x + 4 According to the problem, (x + 4) 2 + x 2 = 20 2 => x 2 + 8x + 16 + x 2 = 400 => 2x 2 + 8x + 16 = 400 => 2x 2 + 8x – 384 = 0 => x 2 + 4x – 192 = 0 => x 2 + 16x – 12x – 192 = 0 => x(x + 16) – 12(x + 16) = 0 => (x + 16)(x – 12) = 0 => x + 16 = 0 and x – 12 = 0 => x = {-16, 12} Since we cannot have a negative measurement in mensuration, the lengths of the sides are 12 and 16
Jennifer jumped off a cliff into the swimming pool. The function h can express her height as a function of time (t) = -16t 2 +16t + 480, where t is the time in seconds and h is the height in feet. a) How long did it take for Jennifer to attain a maximum length. b) What was the highest point that Jennifer reached. c) Calculate the time when Jennifer hit the water?
Comparing the given function with the given function f(x) = ax 2 + bx + c, here a = -16, b = 16, c = 480 a) Finding the vertex will give us the time taken by Jennifer to reach her maximum height x = ${-\dfrac{b}{2a}}$ = ${\dfrac{-16}{2\left( -16\right) }}$ = 0.5 seconds Thus Jennifer took 0.5 seconds to reach her maximum height b) Putting the value of the vertex by substitution in the function, we get ${h\left( \dfrac{1}{2}\right) =-16\left( \dfrac{1}{2}\right) ^{2}+16\left( \dfrac{1}{2}\right) +480}$ => ${-16\left( \dfrac{1}{4}\right) +8+480}$ => 484 feet Thus the highest point that Jennifer reached was 484 feet c) When Jennifer hit the water, her height was 0 Thus, by substituting the value of the height in the function, we get -16t 2 +16t + 480 = 0 => -16(t 2 + t – 30) = 0 => t 2 + t – 30 = 0 => t 2 + 6t – 5t – 30 = 0 => t(t + 6) – 5(t + 6) = 0 => (t + 6)(t – 5) = 0 => t + 6 = 0 or t – 5 = 0 => x = {-6, 5} Since time cannot have any negative value, the time taken by Jennifer to hit the water is 5 seconds.
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Quadratic Equation Solver
We can help you solve an equation of the form " ax 2 + bx + c = 0 " Just enter the values of a, b and c below :
Is it Quadratic?
Only if it can be put in the form ax 2 + bx + c = 0 , and a is not zero .
The name comes from "quad" meaning square, as the variable is squared (in other words x 2 ).
These are all quadratic equations in disguise:
| In disguise | In standard form | a, b and c |
|---|---|---|
| = 3x -1 | - 3x + 1 = 0 | a=1, b=-3, c=1 |
| - 2x) = 5 | - 4x - 5 = 0 | a=2, b=-4, c=-5 |
| - x - 3 = 0 | a=1, b=-1, c=-3 | |
| = 0 | + x - 1 = 0 | a=5, b=1, c=-1 |
How Does this Work?
The solution(s) to a quadratic equation can be calculated using the Quadratic Formula :
The "±" means we need to do a plus AND a minus, so there are normally TWO solutions !
The blue part ( b 2 - 4ac ) is called the "discriminant", because it can "discriminate" between the possible types of answer:
- when it is positive, we get two real solutions,
- when it is zero we get just ONE solution,
- when it is negative we get complex solutions.
Learn more at Quadratic Equations
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Quadratic equation solver This calculator solves quadratic equations using three different methods : the quadratic formula method, completing the square, and the factoring method. Calculator shows all the work and provides detailed explanation on how to solve an equation.
Hire MATHPORTAL experts to do math homework for you. Prices start at $3 per problem. How to use this calculatorThe most commonly used methods for solving quadratic equations are: 1 . Factoring method 2 . Solving quadratic equations by completing the square 3 . Using quadratic formula In the following sections, we'll go over these methods. Method 1A : Factoring methodIf a quadratic trinomial can be factored, this is the best solving method. We often use this method when the leading coefficient is equal to 1 or -1. If this is not the case, then it is better to use some other method. Example 01: Solve x 2 -8x+15=0 by factoring. Here we see that the leading coefficient is 1, so the factoring method is our first choice. To factor this equation, we must find two numbers a and b with a sum of a + b = 8 and a product of a × b = 15 After some trials and errors, we see that a = 3 and b = 5 . Now we use a formula x 2 -8x+15=(x-a)(x-b) to get factored form: x 2 -8x+15=(x-3)(x-5) Divide the factored form into two linear equations to get solutions. Method 1B: Factoring – special casesExample 02: Solve x 2 -8x=0 by factoring. In this case, (when the coefficient c = 0 ) we can factor out x out of x 2 -8 . Example 03: Solve x 2 -16=0 by factoring. In this case, (when the middle term is equal 0) we can use the difference of squares formula. Method 3 : Solve using quadratic formulaThis method solves all types of quadratic equations. It works best when solutions contain some radicals or complex numbers. Example 05: Solve equation $ 2x^2 + 3x - 2 = 0$ by using quadratic formula. Step 1 : Read the values of $a$, $b$, and $c$ from the quadratic equation. ( a is the number in front of x 2 , b is the number in front of x and c is the number at the end) a = 2, b = 3 and c = -2 Step 2 :Plug the values for a, b, and c into the quadratic formula and simplify. Step 3 : Solve for x 1 and x 2 Method 2 : Completing the squareThis method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations. The method involves seven steps. Example 04: Solve equation 2x 2 +8x-10=0 by completing the square. Step 1 : Divide the equation by the number in front of the square term. Step 2 : move -5 to the right: Step 3 : Take half of the x-term coefficient $ \color{blue}{\dfrac{4}{2}} $, square it $ \color{blue}{\left(\dfrac{4}{2} \right)^2} $ and add this value to both sides. Step 4 : Simplify left and right side. x 2 +4x+2 2 =9 Step 5 : Write the perfect square on the left. Step 6 : Take the square root of both sides. Step 7 : Solve for $x_1$ and $x_2$ . 1. Quadratic Equation — step-by-step examples, video tutorials with worked examples. 2. Completing the Square — video on Khan Academy 3. Completing the Square — video on Khan Academy 4. Polynomial equation solver Welcome to MathPortal. This website's owner is mathematician Miloš Petrović. I designed this website and wrote all the calculators, lessons, and formulas . If you want to contact me, probably have some questions, write me using the contact form or email me on [email protected] Email (optional)
Quadratics or Quadratic Equations Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations . The general form of the quadratic equation is: ax² + bx + c = 0 where x is an unknown variable and a, b, c are numerical coefficients. For example, x 2 + 2x +1 is a quadratic or quadratic equation. Here, a ≠ 0 because if it equals zero then the equation will not remain quadratic anymore and it will become a linear equation, such as: Thus, this equation cannot be called a quadratic equation. The terms a, b and c are also called quadratic coefficients. The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations . The roots of any polynomial are the solutions for the given equation. What is Quadratic Equation?The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of: where x is the unknown variable and a, b and c are the constant terms. Standard Form of Quadratic Equation Since the quadratic includes only one unknown term or variable, thus it is called univariate. The power of variable x is always non-negative integers. Hence the equation is a polynomial equation with the highest power as 2. The solution for this equation is the values of x, which are also called zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x on the left-hand side of the equation, it will equal to zero. Therefore, they are called zeros. Quadratics FormulaThe formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be: x = [-b±√(b 2 -4ac)]/2a The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here. Examples of QuadraticsBeneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)
Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.
Following are the examples of a quadratic equation in factored form
Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’
How to Solve Quadratic Equations?There are basically four methods of solving quadratic equations. They are:
Using Quadratic Formula
Factoring of Quadratics
Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors. (2x+3)(x-2)=0 Learn more about the factorization of quadratic equations here. Completing the Square MethodLet us learn this method with example. Example: Solve 2x 2 – x – 1 = 0. First, move the constant term to the other side of the equation. 2x 2 – x = 1 Dividing both sides by 2. x 2 – x/2 = ½ Add the square of half of the coefficient of x, (b/2a) 2 , on both the sides, i.e., 1/16 x 2 – x/2 + 1/16 = ½ + 1/16 Now we can factor the right side, (x-¼) 2 = 9/16 = (¾) 2 Taking root on both sides; X – ¼ = ±3/4 Add ¼ on both sides X = ¼ + ¾ = 4/4 = 1 X = ¼ – ¾ = -2/4 = -½ To learn more about completing the square method, click here . For the given Quadratic equation of the form, ax² + bx + c = 0 Therefore the roots of the given equation can be found by: \(\begin{array}{l}x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\end{array} \) where ± (one plus and one minus) represent two distinct roots of the given equation. Taking the Square RootWe can use this method for the equations such as: x 2 + a 2 = 0 Example: Solve x 2 – 50 = 0. x 2 – 50 = 0 Taking the roots both sides √x 2 = ±√50 x = ±√(2 x 5 x 5) Thus, we got the required solution. Related ArticlesVideo Lesson on Quadratic EquationsRange of quadratic equations.  Solved Problems on Quadratic Equations
Applications of Quadratic EquationsMany real-life word problems can be solved using quadratic equations. While solving word problems, some common quadratic equation applications include speed problems and Geometry area problems.
Example: Find the width of a rectangle of area 336 cm2 if its length is equal to the 4 more than twice its width. Solution: Let x cm be the width of the rectangle. Length = (2x + 4) cm We know that Area of rectangle = Length x Width x(2x + 4) = 336 2x 2 + 4x – 336 = 0 x 2 + 2x – 168 = 0 x 2 + 14x – 12x – 168 = 0 x(x + 14) – 12(x + 14) = 0 (x + 14)(x – 12) = 0 x = -14, x = 12 Measurement cannot be negative. Therefore, Width of the rectangle = x = 12 cm Practice Questions
Frequently Asked Questions on QuadraticsWhat is a quadratic equation, what are the methods to solve a quadratic equation, is x 2 – 1 a quadratic equation, what is the solution of x 2 + 4 = 0, write the quadratic equation in the form of sum and product of roots..
Leave a Comment Cancel replyYour Mobile number and Email id will not be published. Required fields are marked * Request OTP on Voice Call Post My Comment  Thanks a lot ,This was very useful for me x=√9 Squaring both the sides, x^2 = 9 x^2 – 9 = 0 It is a quadratic equation.  Register with BYJU'S & Download Free PDFsRegister with byju's & watch live videos. 
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General Quadratic Word ProblemsProjectiles General Word Problems Max/Min Problems Most quadratic word problems should seem very familiar, as they are built from the linear problems that you've done in the past. You will need to use keywords to interpret the English and, from that, create the quadratic model. Then solve the model for the solutions (that is, the x -intercepts). Content Continues Below MathHelp.com When interpreted — within the context of the exercise — one or both of the solutions to the quadratic equation will provide at least part of the answer they've asked of you. Advertisement
The height is defined in terms of the width, so I'll pick a variable for "width", and then create an expression for the height. Let " w " stand for the width of the picture. The height h is given as being 4/3 of the width, so: h = (4/3) w Then the area, being the product of the width and the height, is given by: = [(4/3) w ][ w ] = (4/3) w 2 = 192 I need to solve this "area" equation for the value of the width, and then back-solve to find the value of the height. (4/3) w 2 = 192 w 2 = 144 w = ± 12 Since I can't have a negative width, I can ignore the " w = −12 " solution. Then the width must be the other solution, 12 , and the height is then: h = (4/3)(12) = 16 I can do a quick check of my answer, since this is a "solving" exercise, by plugging my values back into the original exercise. Since 12 × 16 does indeed equal 192 , my answer checks. So my hand-in answer is: The enlargement will be 12 inches by 16 inches.
Remember that consecutive integers are one unit apart, so my numbers are n and n + 1 . Multiplying to get the product, I get: n ( n + 1) = 1122 n 2 + n = 1122 n 2 + n − 1122 = 0 ( n + 34)( n − 33) = 0 The solutions are n = −34 and n = 33 . I need a negative value (because the exercise statement specified that the numbers are negative), so I'll ignore the n = 33 & and take n = −34 . Then the other number is: n + 1 = (−34) + 1 = −33 Then my hand-in answer is: The two numbers are −33 and −34 . Note that the second value could have been gotten by changing the sign on the extraneous solution. Warning: Many students get in the very bad habit of arbitrarily changing signs to get the answers they need, but this does not always work, and will very likely get them in trouble later on. (And it'll cause trouble right now, if the grader is paying attention.) Take the extra half a second to find the right answer the right way.
The first thing I need to do is draw a picture. Since I don't know how wide the path will be, I'll label the width as " x ". Looking at my picture, I see that the total width of the finished garden-plus-path area will be: x + 12 + x = 12 + 2 x ...and the total length will be: x + 16 + x = 16 + 2 x Then the new area, being the product of the new width and height, is given by: (12 + 2 x )(16 + 2 x ) = 285 192 + 56 x + 4 x 2 = 285 4 x 2 + 56 x − 93 = 0 This quadratic is messy enough that I won't bother with trying to use factoring to solve; I'll just go straight to the Quadratic Formula : Obviously the negative value won't work in this context, so I'll ignore it. Checking the original exercise to verify what I'm being asked to find, I notice that I need to have units on my answer: The width of the pathway will be 1.5 meters. 
When dealing with geometric sorts of word problems, it is usually helpful to draw a picture. Since I'll be cutting equal-sized squares out of all of the corners, and since the box will have a square bottom, I know I'll be starting with a square piece of cardboard. I don't know how big the cardboard will be yet, so I'll label the sides as having length " w ". Since I know I'll be cutting out three-by-three squares to get sides that are three inches high, I can mark that on my drawing. The dashed (rather than solid) lines show where I'll be scoring the cardboard and folding up the sides. Since I'll be losing three inches on either end of the cardboard when I fold up the sides, the final width of the bottom will be the original " w " inches, less three on the one side and another three on the other side. That is, the width of the bottom will be: w − 3 − 3 = w − 6 The volume will be the product of the width, the length (which is the same as the width), and the depth (which was created by cutting out the corners and folding up the sides). Then the volume of the box, working from the drawing, gives me the following equation: ( w − 6)( w − 6)(3) = 42 ( w − 6)( w − 6) = 14 ( w − 6) 2 = 14 This is the quadratic I need to solve. I can take the square root of either side, and then add the to the right-hand side: ...or I can multiply out the square and apply the Quadratic Formula : w 2 − 12 w + 36 = 14 w 2 − 12 w + 22 = 0 Either way, I get two solutions which, when expressed in practical decimal terms, tell me that the width of the original cardboard is either about 2.26 inches or else about 9.74 inches. How do I know which solution value for the width is right? By checking each value in the original word problem. If the cardboard is only 2.26 inches wide, then how on earth would I be able to fold up three-inch-deep sides? But if the cardboard is 9.74 inches, then I can fold up three inches of cardboard on either side, and still be left with 3.74 inches in the middle. Checking: (3.74)(3.74)(3) = 41.9628 This isn't exactly 42 , but, taking round-off error into account, it's close enough that I can trust that I have the correct value. So my hand-in answer, complete with units, is: The cardboard should measure 9.75 inches on a side. In this last exercise above, you should notice that each solution method (factoring and the Quadratic Formula) gave the same final answer for the cardboard's width. But the Quadratic Formula took longer and provided me with more opportunities to make mistakes. Moral? Don't get stuck in the rut of always using the Quadratic Formula. URL: https://www.purplemath.com/modules/quadprob2.htm Page 1 Page 2 Page 3 Standardized Test PrepCollege math, homeschool math, share this page.
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How to Solve Quadratics with Complex NumbersQuadratic equations are of the form: ax 2 + bx + c = 0 Where a, b, and c are constants. When solving these equations, the solutions (roots) can be real or complex numbers. Complex roots occur when the discriminant b 2 −4ac is negative. Steps to to Solve Quadratics with Complex NumbersStep 1: Ensure the quadratic equation is in standard form: Step 2: Calculate the Discriminant The discriminant (Δ) is calculated as: Δ = b 2 −4ac Step 3: Determine the Nature of the Roots
Since we are focusing on complex roots, we consider Δ < 0. Step 4: Use the Quadratic Formula For complex roots, use the quadratic formula: x = [Tex]\frac{-b \pm \sqrt{\Delta}}{2a}[/Tex] Since Δ is negative, Δ will be an imaginary number. Let’s rewrite the discriminant: [Tex]\Delta = b^2 – 4ac = -k \quad \text{where} \quad k > 0[/Tex] Thus, [Tex]\sqrt{\Delta} = \sqrt{-k} = i\sqrt{k}[/Tex] Step 5: Substitute and Simplify Substitute Δ = i√k into the quadratic formula: [Tex]x = \frac{-b \pm i\sqrt{k}}{2a}[/Tex] This yields the complex roots: [Tex]x_1 = \frac{-b + i\sqrt{k}}{2a}[/Tex] and [Tex] x_2 = \frac{-b – i\sqrt{k}}{2a}[/Tex] Example of Quadratics with Complex NumbersExample: Solve the quadratic equation x 2 + 4x + 13 = 0. Write in standard form : x 2 + 4x + 13 = 0 Calculate the discriminant : Δ = 4 2 − 4 ⋅ 1 ⋅ 13 = 16 − 52 = −36 Determine the nature of the roots : Since Δ < 0, the roots are complex. Use the quadratic formula : x = [− 4 ± √(−36)]/(2 ⋅ 1) = (−4 ± 6i)/2 Simplify : x 1 = (−4 + 6i)/2 = −2 + 3i and x 2 = (−4 – 6i)/2 = −2 – 3i Thus, the solutions are x 1 = −2 + 3i and x 2 = −2 − 3i. When the discriminant of a quadratic equation is negative, the roots are complex. By using the quadratic formula and simplifying the imaginary parts, we can find the complex roots of the quadratic equation.
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 Trigonometric Equations - Multiple-Choice WS/Practice (20 problems + solutions)Subject: Mathematics Age range: 16+ Resource type: Worksheet/Activity  Last updated 18 August 2024
This is an engaging multiple-choice practice on solving trigonometric equations(basic, multiple angles, of quadratic type, such that can be solved by factoring). There are 5 pages, containing a total of 20 problems each having five optional answers. The first 8 problems are finding all the solutions of each equation given, the next 12 problems are determining the sum of the solutions of an equation on a closed interval. Students have an empty field below each problem where to write down their solution. Detailed typed solutions are provided. Tes paid licence How can I reuse this? Your rating is required to reflect your happiness. It's good to leave some feedback. Something went wrong, please try again later. This resource hasn't been reviewed yet To ensure quality for our reviews, only customers who have purchased this resource can review it Report this resource to let us know if it violates our terms and conditions. Our customer service team will review your report and will be in touch. Not quite what you were looking for? Search by keyword to find the right resource:Information
InitiativesYou are accessing a machine-readable page. In order to be human-readable, please install an RSS reader. All articles published by MDPI are made immediately available worldwide under an open access license. No special permission is required to reuse all or part of the article published by MDPI, including figures and tables. For articles published under an open access Creative Common CC BY license, any part of the article may be reused without permission provided that the original article is clearly cited. For more information, please refer to https://www.mdpi.com/openaccess . Feature papers represent the most advanced research with significant potential for high impact in the field. A Feature Paper should be a substantial original Article that involves several techniques or approaches, provides an outlook for future research directions and describes possible research applications. Feature papers are submitted upon individual invitation or recommendation by the scientific editors and must receive positive feedback from the reviewers. Editor’s Choice articles are based on recommendations by the scientific editors of MDPI journals from around the world. Editors select a small number of articles recently published in the journal that they believe will be particularly interesting to readers, or important in the respective research area. The aim is to provide a snapshot of some of the most exciting work published in the various research areas of the journal. Original Submission Date Received: .
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Find support for a specific problem in the support section of our website. Please let us know what you think of our products and services. Visit our dedicated information section to learn more about MDPI. JSmol ViewerDesign of diffractive neural networks for solving different classification problems at different wavelengths.  1. Introduction2. design of spectral dnns for solving several classification problems, 3. gradient method for designing spectral dnns, 4. design examples of spectral dnns, 4.1. sequential solution of the classification problems, 4.2. parallel solution of the classification problems, 5. discussion and conclusions, supplementary materials, author contributions, institutional review board statement, informed consent statement, data availability statement, conflicts of interest.
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Share and CiteMotz, G.A.; Doskolovich, L.L.; Soshnikov, D.V.; Byzov, E.V.; Bezus, E.A.; Golovastikov, N.V.; Bykov, D.A. Design of Diffractive Neural Networks for Solving Different Classification Problems at Different Wavelengths. Photonics 2024 , 11 , 780. https://doi.org/10.3390/photonics11080780 Motz GA, Doskolovich LL, Soshnikov DV, Byzov EV, Bezus EA, Golovastikov NV, Bykov DA. Design of Diffractive Neural Networks for Solving Different Classification Problems at Different Wavelengths. Photonics . 2024; 11(8):780. https://doi.org/10.3390/photonics11080780 Motz, Georgy A., Leonid L. Doskolovich, Daniil V. Soshnikov, Egor V. Byzov, Evgeni A. Bezus, Nikita V. Golovastikov, and Dmitry A. Bykov. 2024. "Design of Diffractive Neural Networks for Solving Different Classification Problems at Different Wavelengths" Photonics 11, no. 8: 780. https://doi.org/10.3390/photonics11080780 Article MetricsSupplementary material. ZIP-Document (ZIP, 337 KiB) Further InformationMdpi initiatives, follow mdpi.  Subscribe to receive issue release notifications and newsletters from MDPI journals | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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Use a Problem-Solving Strategy. Read the problem. Make sure all the words and ideas are understood. Identify what we are looking for.; Name what we are looking for. Choose a variable to represent that quantity. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information.
To solve a quadratic equation, use the quadratic formula: x = (-b ± √ (b^2 - 4ac)) / (2a). There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution. In math, a quadratic equation is a second-order polynomial equation in ...
This is how the solution of the equation 2 x 2 − 12 x + 18 = 0 goes: 2 x 2 − 12 x + 18 = 0 x 2 − 6 x + 9 = 0 Divide by 2. ( x − 3) 2 = 0 Factor. ↓ x − 3 = 0 x = 3. All terms originally had a common factor of 2 , so we divided all sides by 2 —the zero side remained zero—which made the factorization easier.
Test your understanding of Quadratic equations & functions with these NaN questions. Start test. This topic covers: - Solving quadratic equations - Graphing quadratic functions - Features of quadratic functions - Quadratic equations/functions word problems - Systems of quadratic equations - Quadratic inequalities.
Solve by completing the square: Non-integer solutions. Worked example: completing the square (leading coefficient ≠ 1) Solving quadratics by completing the square: no solution. Proof of the quadratic formula. Solving quadratics by completing the square. Completing the square review. Quadratic formula proof review.
A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. ... among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires ...
To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x.
There are different methods you can use to solve quadratic equations, depending on your particular problem. Solve By Factoring. Example: 3x^2-2x-1=0. Complete The Square. Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.) Take the Square Root. Example: 2x^2=18. Quadratic Formula. Example ...
First, we need to rewrite the given quadratic equation in Standard Form, After getting the correct standard form in the previous step, it's now time to plug the values of. From the converted standard form, extract the required values. Then evaluate these values into the quadratic formula. Solving Quadratic Equations by Completing the Square.
Quadratic Equation in Standard Form: ax 2 + bx + c = 0. Quadratic Equations can be factored. Quadratic Formula: x = −b ± √ (b2 − 4ac) 2a. When the Discriminant ( b2−4ac) is: positive, there are 2 real solutions. zero, there is one real solution. negative, there are 2 complex solutions.
An expression of the form ax 2 + bx + c, where a ≠ 0 is called a quadratic expression. In other words, any expression with the variables of the highest exponent or the expression's degree as 2 is a quadratic expression The standard form of a quadratic expression looks like this. Here are some examples of expressions: 1. 2.
Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them: By Completing the Square; By Factoring; By Quadratic Formula; By graphing; For each process, follow the following typical steps: Make the equation; Solve for the unknown variable using the appropriate method; Interpret the result
A Quadratic Equation! Let us solve it using our Quadratic Equation Solver. Enter 1, −1 and −6; And you should get the answers −2 and 3; R 1 cannot be negative, so R 1 = 3 Ohms is the answer. The two resistors are 3 ohms and 6 ohms. Others. Quadratic Equations are useful in many other areas: For a parabolic mirror, a reflecting telescope ...
Methods to Solve Quadratic Equations: Factoring; Square Root Property; Completing the Square; Quadratic Formula; How to identify the most appropriate method to solve a quadratic equation. Try Factoring first. If the quadratic factors easily, this method is very quick. Try the Square Root Property next. If the equation fits the form \(a x^{2}=k ...
The solution (s) to a quadratic equation can be calculated using the Quadratic Formula: The "±" means we need to do a plus AND a minus, so there are normally TWO solutions ! The blue part ( b2 - 4ac) is called the "discriminant", because it can "discriminate" between the possible types of answer: when it is negative we get complex solutions.
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.
There are many ways to solve quadratic equations, such as through factoring, completing the square, or using the quadratic formula. In the following section, we will demonstrate using the quadratic formula. ... Start with our practice problems at the top of this page and see if you can solve them. If you run into trouble, ...
Example 05: Solve equation 2x2 + 3x − 2 = 0 by using quadratic formula. Step 1: Read the values of a, b, and c from the quadratic equation. ( a is the number in front of x 2 , b is the number in front of x and c is the number at the end) Step 2 :Plug the values for a, b, and c into the quadratic formula and simplify.
Quadratics or Quadratic Equations. Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations. The general form of the quadratic equation is: ax² + bx + c = 0. where x is an unknown variable and a, b, c are numerical ...
Most quadratic word problems should seem very familiar, as they are built from the linear problems that you've done in the past. You will need to use keywords to interpret the English and, from that, create the quadratic model. Then solve the model for the solutions (that is, the x-intercepts).
High School Math Solutions - Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c... Enter a problem
In mathematics, a quadratic equation (from Latin quadratus 'square') is an equation that can be rearranged in standard form as [1] + + =, where x represents an unknown value, and a, b, and c represent known numbers, where a ≠ 0. (If a = 0 and b ≠ 0 then the equation is linear, not quadratic.)The numbers a, b, and c are the coefficients of the equation and may be distinguished by ...
Worked example. First we need to identify the values for a, b, and c (the coefficients). First step, make sure the equation is in the format from above, a x 2 + b x + c = 0 : is what makes it a quadratic). Then we plug a , b , and c into the formula: solving this looks like: Therefore x = 3 or x = − 7 .
This coloring activity provides students with 12 problems. Students will solve quadratic equations by factoring. Six of the equations have the leading coefficient equal to 1 and the other half have the leading coefficient >1. After solving a problem, the students find their answer in a table. This tells them what color to use in the coloring page.
Quadratic equations are of the form: ax 2 + bx + c = 0. Where a, b, and c are constants. When solving these equations, the solutions (roots) can be real or complex numbers. Complex roots occur when the discriminant b 2 −4ac is negative. Steps to to Solve Quadratics with Complex Numbers. Step 1: Ensure the quadratic equation is in standard form:
Text: Elementary Differential Equations & Boundary Value Problems, 12th Edition, Boyce, DiPrima, and Meade, Wiley and Sons, Inc. (2022) ... how to solve partial differential equations by separation of variables or Fourier series. to apply these techniques to the three classical equations: the heat, wave, and Laplace's equation ...
There are 5 pages, containing a total of 20 problems each having five optional answers. The first 8 problems are finding all the solutions of each equation given, the next 12 problems are determining the sum of the solutions of an equation on a closed interval. Students have an empty field below each problem where to write down their solution.
We consider the problem of designing a diffractive neural network (DNN) consisting of a set of sequentially placed phase diffractive optical elements (DOEs) and intended for the optical solution of several given classification problems at different operating wavelengths, so that each classification problem is solved at the corresponding wavelength. The problem of calculating the DNN is ...