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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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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

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Quadratic Equations and Functions

Solve Quadratic Equations Using the Quadratic Formula

Learning Objectives

By the end of this section, you will be able to:

• Solve quadratic equations using the Quadratic Formula
• Use the discriminant to predict the number and type of solutions of a quadratic equation
• Identify the most appropriate method to use to solve a quadratic equation

Before you get started, take this readiness quiz.

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for x.

We start with the standard form of a quadratic equation and solve it for x by completing the square.

To use the Quadratic Formula , we substitute the values of a , b , and c from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.

Notice the formula is an equation. Make sure you use both sides of the equation.

• Write the quadratic equation in standard form, ax 2 + bx + c = 0. Identify the values of a , b , and c .
• Write the Quadratic Formula. Then substitute in the values of a , b , and c .
• Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “ x =”.

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula . If we get a radical as a solution, the final answer must have the radical in its simplified form.

When we substitute a , b , and c into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

Remember, to use the Quadratic Formula, the equation must be written in standard form, ax 2 + bx + c = 0. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Our first step is to get the equation in standard form.

When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.

Our first step is to clear the fractions.

Think about the equation ( x − 3) 2 = 0. We know from the Zero Product Property that this equation has only one solution,

We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to 0 gives just one solution. Notice that once the radicand is simplified it becomes 0 , which leads to only one solution.

Did you recognize that 4 x 2 − 20 x + 25 is a perfect square trinomial. It is equivalent to (2 x − 5) 2 ? If you solve

Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant .

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.

• If b 2 − 4 ac > 0, the equation has 2 real solutions.
• if b 2 − 4 ac = 0, the equation has 1 real solution.
• if b 2 − 4 ac < 0, the equation has 2 complex solutions.

Determine the number of solutions to each quadratic equation.

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

Since the discriminant is positive, there are 2 real solutions to the equation.

Since the discriminant is negative, there are 2 complex solutions to the equation.

Since the discriminant is 0, there is 1 real solution to the equation.

Determine the numberand type of solutions to each quadratic equation.

ⓐ 2 complex solutions; ⓑ 2 real solutions; ⓒ 1 real solution

Determine the number and type of solutions to each quadratic equation.

ⓐ 2 real solutions; ⓑ 2 complex solutions; ⓒ 1 real solution

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We summarize the four methods that we have used to solve quadratic equations below.

• Square Root Property
• Completing the Square
• Quadratic Formula

What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.

• Try Factoring first. If the quadratic factors easily, this method is very quick.

• Use the Quadratic Formula . Any other quadratic equation is best solved by using the Quadratic Formula.

The next example uses this strategy to decide how to solve each quadratic equation.

Identify the most appropriate method to use to solve each quadratic equation.

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.

ⓐ factoring; ⓑ Square Root Property; ⓒ Quadratic Formula

ⓐ Quadratic Forumula;

Access these online resources for additional instruction and practice with using the Quadratic Formula.

• Using the Quadratic Formula
• Solve a Quadratic Equation Using the Quadratic Formula with Complex Solutions
• Discriminant in Quadratic Formula

Key Concepts

• Write the quadratic equation in standard form, ax 2 + bx + c = 0. Identify the values of a , b , c .
• Write the Quadratic Formula. Then substitute in the values of a , b , c .
• Try the Square Root Property next. If the equation fits the form ax 2 = k or a ( x − h ) 2 = k , it can easily be solved by using the Square Root Property.
• Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.

Practice Makes Perfect

In the following exercises, solve by using the Quadratic Formula.

Use the Discriminant to Predict the Number of Real Solutions of a Quadratic Equation

In the following exercises, determine the number of real solutions for each quadratic equation.

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

Writing Exercises

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?

Answers will vary.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Intermediate Algebra Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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10.3 Solve Quadratic Equations Using the Quadratic Formula

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic equations using the quadratic formula
• Use the discriminant to predict the number of solutions of a quadratic equation
• Identify the most appropriate method to use to solve a quadratic equation

Be Prepared 10.7

Before you get started, take this readiness quiz.

Simplify: −20 − 5 10 −20 − 5 10 . If you missed this problem, review Example 1.74 .

Be Prepared 10.8

Simplify: 4 + 121 4 + 121 . If you missed this problem, review Example 9.29 .

Be Prepared 10.9

Simplify: 128 128 . If you missed this problem, review Example 9.12 .

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes.’ In this section, we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’ so that we would do the algebraic steps only once and then use the new formula to find the value of the specific variable. Now, we will go through the steps of completing the square in general to solve a quadratic equation for x . It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

This last equation is the Quadratic Formula.

• Quadratic Formula

The solutions to a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 , a ≠ 0 a ≠ 0 are given by the formula:

To use the Quadratic Formula, we substitute the values of a , b , and c a , b , and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

Example 10.28

How to solve a quadratic equation using the quadratic formula.

Solve 2 x 2 + 9 x − 5 = 0 2 x 2 + 9 x − 5 = 0 by using the Quadratic Formula.

Try It 10.55

Solve 3 y 2 − 5 y + 2 = 0 3 y 2 − 5 y + 2 = 0 by using the Quadratic Formula.

Try It 10.56

Solve 4 z 2 + 2 z − 6 = 0 4 z 2 + 2 z − 6 = 0 by using the Quadratic Formula.

Solve a quadratic equation using the Quadratic Formula.

• Step 1. Write the Quadratic Formula in standard form. Identify the a a , b b , and c c values.
• Step 2. Write the Quadratic Formula. Then substitute in the values of a a , b b , and c . c .
• Step 3. Simplify.
• Step 4. Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time. And remember, the Quadratic Formula is an equation. Be sure you start with ‘ x = x = ’.

Example 10.29

Solve x 2 − 6 x + 5 = 0 x 2 − 6 x + 5 = 0 by using the Quadratic Formula.

Try It 10.57

Solve a 2 − 2 a − 15 = 0 a 2 − 2 a − 15 = 0 by using the Quadratic Formula.

Try It 10.58

Solve b 2 + 10 b + 24 = 0 b 2 + 10 b + 24 = 0 by using the Quadratic Formula.

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula. If we get a radical as a solution, the final answer must have the radical in its simplified form.

Example 10.30

Solve 4 y 2 − 5 y − 3 = 0 4 y 2 − 5 y − 3 = 0 by using the Quadratic Formula.

We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named ‘ x ’.

Try It 10.59

Solve 2 p 2 + 8 p + 5 = 0 2 p 2 + 8 p + 5 = 0 by using the Quadratic Formula.

Try It 10.60

Solve 5 q 2 − 11 q + 3 = 0 5 q 2 − 11 q + 3 = 0 by using the Quadratic Formula.

Example 10.31

Solve 2 x 2 + 10 x + 11 = 0 2 x 2 + 10 x + 11 = 0 by using the Quadratic Formula.

Try It 10.61

Solve 3 m 2 + 12 m + 7 = 0 3 m 2 + 12 m + 7 = 0 by using the Quadratic Formula.

Try It 10.62

Solve 5 n 2 + 4 n − 4 = 0 5 n 2 + 4 n − 4 = 0 by using the Quadratic Formula.

We cannot take the square root of a negative number. So, when we substitute a a , b b , and c c into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example.

Example 10.32

Solve 3 p 2 + 2 p + 9 = 0 3 p 2 + 2 p + 9 = 0 by using the Quadratic Formula.

Try It 10.63

Solve 4 a 2 − 3 a + 8 = 0 4 a 2 − 3 a + 8 = 0 by using the Quadratic Formula.

Try It 10.64

Solve 5 b 2 + 2 b + 4 = 0 5 b 2 + 2 b + 4 = 0 by using the Quadratic Formula.

The quadratic equations we have solved so far in this section were all written in standard form, a x 2 + b x + c = 0 a x 2 + b x + c = 0 . Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example 10.33

Solve x ( x + 6 ) + 4 = 0 x ( x + 6 ) + 4 = 0 by using the Quadratic Formula.

Try It 10.65

Solve x ( x + 2 ) − 5 = 0 x ( x + 2 ) − 5 = 0 by using the Quadratic Formula.

Try It 10.66

Solve y ( 3 y − 1 ) − 2 = 0 y ( 3 y − 1 ) − 2 = 0 by using the Quadratic Formula.

When we solved linear equations, if an equation had too many fractions we ‘cleared the fractions’ by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions—to solve. We can use the same strategy with quadratic equations.

Example 10.34

Solve 1 2 u 2 + 2 3 u = 1 3 1 2 u 2 + 2 3 u = 1 3 by using the Quadratic Formula.

Try It 10.67

Solve 1 4 c 2 − 1 3 c = 1 12 1 4 c 2 − 1 3 c = 1 12 by using the Quadratic Formula.

Try It 10.68

Solve 1 9 d 2 − 1 2 d = − 1 2 1 9 d 2 − 1 2 d = − 1 2 by using the Quadratic Formula.

Think about the equation ( x − 3 ) 2 = 0 ( x − 3 ) 2 = 0 . We know from the Zero Products Principle that this equation has only one solution: x = 3 x = 3 .

We will see in the next example how using the Quadratic Formula to solve an equation with a perfect square also gives just one solution.

Example 10.35

Solve 4 x 2 − 20 x = −25 4 x 2 − 20 x = −25 by using the Quadratic Formula.

Did you recognize that 4 x 2 − 20 x + 25 4 x 2 − 20 x + 25 is a perfect square?

Try It 10.69

Solve r 2 + 10 r + 25 = 0 r 2 + 10 r + 25 = 0 by using the Quadratic Formula.

Try It 10.70

Solve 25 t 2 − 40 t = −16 25 t 2 − 40 t = −16 by using the Quadratic Formula.

Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation?

Yes, the quantity inside the radical of the Quadratic Formula makes it easy for us to determine the number of solutions. This quantity is called the discriminant .

Discriminant

In the Quadratic Formula x = − b ± b 2 − 4 a c 2 a x = − b ± b 2 − 4 a c 2 a , the quantity b 2 − 4 a c b 2 − 4 a c is called the discriminant .

Let’s look at the discriminant of the equations in Example 10.28 , Example 10.32 , and Example 10.35 , and the number of solutions to those quadratic equations.

When the discriminant is positive ( x = − b ± + 2 a ) ( x = − b ± + 2 a ) the quadratic equation has two solutions .

When the discriminant is zero ( x = − b ± 0 2 a ) ( x = − b ± 0 2 a ) the quadratic equation has one solution .

When the discriminant is negative ( x = − b ± − 2 a ) ( x = − b ± − 2 a ) the quadratic equation has no real solutions .

Use the discriminant, b 2 − 4 a c b 2 − 4 a c , to determine the number of solutions of a Quadratic Equation.

For a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 , a ≠ 0 a ≠ 0 ,

• if b 2 − 4 a c > 0 b 2 − 4 a c > 0 , the equation has two solutions.
• if b 2 − 4 a c = 0 b 2 − 4 a c = 0 , the equation has one solution.
• if b 2 − 4 a c < 0 b 2 − 4 a c < 0 , the equation has no real solutions.

Example 10.36

Determine the number of solutions to each quadratic equation:

ⓐ 2 v 2 − 3 v + 6 = 0 2 v 2 − 3 v + 6 = 0 ⓑ 3 x 2 + 7 x − 9 = 0 3 x 2 + 7 x − 9 = 0 ⓒ 5 n 2 + n + 4 = 0 5 n 2 + n + 4 = 0 ⓓ 9 y 2 − 6 y + 1 = 0 9 y 2 − 6 y + 1 = 0

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

Try It 10.71

ⓐ 8 m 2 − 3 m + 6 = 0 8 m 2 − 3 m + 6 = 0 ⓑ 5 z 2 + 6 z − 2 = 0 5 z 2 + 6 z − 2 = 0 ⓒ 9 w 2 + 24 w + 16 = 0 9 w 2 + 24 w + 16 = 0 ⓓ 9 u 2 − 2 u + 4 = 0 9 u 2 − 2 u + 4 = 0

Try It 10.72

ⓐ b 2 + 7 b − 13 = 0 b 2 + 7 b − 13 = 0 ⓑ 5 a 2 − 6 a + 10 = 0 5 a 2 − 6 a + 10 = 0 ⓒ 4 r 2 − 20 r + 25 = 0 4 r 2 − 20 r + 25 = 0 ⓓ 7 t 2 − 11 t + 3 = 0 7 t 2 − 11 t + 3 = 0

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We have used four methods to solve quadratic equations:

• Square Root Property
• Completing the Square

You can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method to use.

Identify the most appropriate method to solve a Quadratic Equation.

• Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
• Step 2. Try the Square Root Property next. If the equation fits the form a x 2 = k a x 2 = k or a ( x − h ) 2 = k a ( x − h ) 2 = k , it can easily be solved by using the Square Root Property.
• Step 3. Use the Quadratic Formula . Any quadratic equation can be solved by using the Quadratic Formula.

What about the method of completing the square? Most people find that method cumbersome and prefer not to use it. We needed to include it in this chapter because we completed the square in general to derive the Quadratic Formula. You will also use the process of completing the square in other areas of algebra.

Example 10.37

Identify the most appropriate method to use to solve each quadratic equation:

ⓐ 5 z 2 = 17 5 z 2 = 17 ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0 ⓒ 8 u 2 + 6 u = 11 8 u 2 + 6 u = 11

ⓐ 5 z 2 = 17 5 z 2 = 17

Since the equation is in the a x 2 = k a x 2 = k , the most appropriate method is to use the Square Root Property.

ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0

We recognize that the left side of the equation is a perfect square trinomial, and so Factoring will be the most appropriate method.

ⓒ 8 u 2 + 6 u = 11 8 u 2 + 6 u = 11

Put the equation in standard form. 8 u 2 + 6 u − 11 = 0 8 u 2 + 6 u − 11 = 0

While our first thought may be to try Factoring, thinking about all the possibilities for trial and error leads us to choose the Quadratic Formula as the most appropriate method

Try It 10.73

ⓐ x 2 + 6 x + 8 = 0 x 2 + 6 x + 8 = 0 ⓑ ( n − 3 ) 2 = 16 ( n − 3 ) 2 = 16 ⓒ 5 p 2 − 6 p = 9 5 p 2 − 6 p = 9

Try It 10.74

ⓐ 8 a 2 + 3 a − 9 = 0 8 a 2 + 3 a − 9 = 0 ⓑ 4 b 2 + 4 b + 1 = 0 4 b 2 + 4 b + 1 = 0 ⓒ 5 c 2 = 125 5 c 2 = 125

Access these online resources for additional instruction and practice with using the Quadratic Formula:

• Solving Quadratic Equations: Solving with the Quadratic Formula
• How to solve a quadratic equation in standard form using the Quadratic Formula (example)
• Solving Quadratic Equations using the Quadratic Formula—Example 3
• Solve Quadratic Equations using Quadratic Formula

Section 10.3 Exercises

Practice makes perfect.

Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

4 m 2 + m − 3 = 0 4 m 2 + m − 3 = 0

4 n 2 − 9 n + 5 = 0 4 n 2 − 9 n + 5 = 0

2 p 2 − 7 p + 3 = 0 2 p 2 − 7 p + 3 = 0

3 q 2 + 8 q − 3 = 0 3 q 2 + 8 q − 3 = 0

p 2 + 7 p + 12 = 0 p 2 + 7 p + 12 = 0

q 2 + 3 q − 18 = 0 q 2 + 3 q − 18 = 0

r 2 − 8 r − 33 = 0 r 2 − 8 r − 33 = 0

t 2 + 13 t + 40 = 0 t 2 + 13 t + 40 = 0

3 u 2 + 7 u − 2 = 0 3 u 2 + 7 u − 2 = 0

6 z 2 − 9 z + 1 = 0 6 z 2 − 9 z + 1 = 0

2 a 2 − 6 a + 3 = 0 2 a 2 − 6 a + 3 = 0

5 b 2 + 2 b − 4 = 0 5 b 2 + 2 b − 4 = 0

2 x 2 + 3 x + 9 = 0 2 x 2 + 3 x + 9 = 0

6 y 2 − 5 y + 2 = 0 6 y 2 − 5 y + 2 = 0

v ( v + 5 ) − 10 = 0 v ( v + 5 ) − 10 = 0

3 w ( w − 2 ) − 8 = 0 3 w ( w − 2 ) − 8 = 0

1 3 m 2 + 1 12 m = 1 4 1 3 m 2 + 1 12 m = 1 4

1 3 n 2 + n = − 1 2 1 3 n 2 + n = − 1 2

16 c 2 + 24 c + 9 = 0 16 c 2 + 24 c + 9 = 0

25 d 2 − 60 d + 36 = 0 25 d 2 − 60 d + 36 = 0

5 m 2 + 2 m − 7 = 0 5 m 2 + 2 m − 7 = 0

8 n 2 − 3 n + 3 = 0 8 n 2 − 3 n + 3 = 0

p 2 − 6 p − 27 = 0 p 2 − 6 p − 27 = 0

25 q 2 + 30 q + 9 = 0 25 q 2 + 30 q + 9 = 0

4 r 2 + 3 r − 5 = 0 4 r 2 + 3 r − 5 = 0

3 t ( t − 2 ) = 2 3 t ( t − 2 ) = 2

2 a 2 + 12 a + 5 = 0 2 a 2 + 12 a + 5 = 0

4 d 2 − 7 d + 2 = 0 4 d 2 − 7 d + 2 = 0

3 4 b 2 + 1 2 b = 3 8 3 4 b 2 + 1 2 b = 3 8

1 9 c 2 + 2 3 c = 3 1 9 c 2 + 2 3 c = 3

2 x 2 + 12 x − 3 = 0 2 x 2 + 12 x − 3 = 0

16 y 2 + 8 y + 1 = 0 16 y 2 + 8 y + 1 = 0

In the following exercises, determine the number of solutions to each quadratic equation.

• ⓐ 4 x 2 − 5 x + 16 = 0 4 x 2 − 5 x + 16 = 0
• ⓑ 36 y 2 + 36 y + 9 = 0 36 y 2 + 36 y + 9 = 0
• ⓒ 6 m 2 + 3 m − 5 = 0 6 m 2 + 3 m − 5 = 0
• ⓓ 18 n 2 − 7 n + 3 = 0 18 n 2 − 7 n + 3 = 0
• ⓐ 9 v 2 − 15 v + 25 = 0 9 v 2 − 15 v + 25 = 0
• ⓑ 100 w 2 + 60 w + 9 = 0 100 w 2 + 60 w + 9 = 0
• ⓒ 5 c 2 + 7 c − 10 = 0 5 c 2 + 7 c − 10 = 0
• ⓓ 15 d 2 − 4 d + 8 = 0 15 d 2 − 4 d + 8 = 0
• ⓐ r 2 + 12 r + 36 = 0 r 2 + 12 r + 36 = 0
• ⓑ 8 t 2 − 11 t + 5 = 0 8 t 2 − 11 t + 5 = 0
• ⓒ 4 u 2 − 12 u + 9 = 0 4 u 2 − 12 u + 9 = 0
• ⓓ 3 v 2 − 5 v − 1 = 0 3 v 2 − 5 v − 1 = 0
• ⓐ 25 p 2 + 10 p + 1 = 0 25 p 2 + 10 p + 1 = 0
• ⓑ 7 q 2 − 3 q − 6 = 0 7 q 2 − 3 q − 6 = 0
• ⓒ 7 y 2 + 2 y + 8 = 0 7 y 2 + 2 y + 8 = 0
• ⓓ 25 z 2 − 60 z + 36 = 0 25 z 2 − 60 z + 36 = 0

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

ⓐ x 2 − 5 x − 24 = 0 x 2 − 5 x − 24 = 0 ⓑ ( y + 5 ) 2 = 12 ( y + 5 ) 2 = 12 ⓒ 14 m 2 + 3 m = 11 14 m 2 + 3 m = 11

ⓐ ( 8 v + 3 ) 2 = 81 ( 8 v + 3 ) 2 = 81 ⓑ w 2 − 9 w − 22 = 0 w 2 − 9 w − 22 = 0 ⓒ 4 n 2 − 10 = 6 4 n 2 − 10 = 6

ⓐ 6 a 2 + 14 = 20 6 a 2 + 14 = 20 ⓑ ( x − 1 4 ) 2 = 5 16 ( x − 1 4 ) 2 = 5 16 ⓒ y 2 − 2 y = 8 y 2 − 2 y = 8

ⓐ 8 b 2 + 15 b = 4 8 b 2 + 15 b = 4 ⓑ 5 9 v 2 − 2 3 v = 1 5 9 v 2 − 2 3 v = 1 ⓒ ( w + 4 3 ) 2 = 2 9 ( w + 4 3 ) 2 = 2 9

Everyday Math

A flare is fired straight up from a ship at sea. Solve the equation 16 ( t 2 − 13 t + 40 ) = 0 16 ( t 2 − 13 t + 40 ) = 0 for t t , the number of seconds it will take for the flare to be at an altitude of 640 feet.

An architect is designing a hotel lobby. She wants to have a triangular window looking out to an atrium, with the width of the window 6 feet more than the height. Due to energy restrictions, the area of the window must be 140 square feet. Solve the equation 1 2 h 2 + 3 h = 140 1 2 h 2 + 3 h = 140 for h h , the height of the window.

Writing Exercises

Solve the equation x 2 + 10 x = 200 x 2 + 10 x = 200 ⓐ by completing the square ⓑ using the Quadratic Formula ⓒ Which method do you prefer? Why?

Solve the equation 12 y 2 + 23 y = 24 12 y 2 + 23 y = 24 ⓐ by completing the square ⓑ using the Quadratic Formula ⓒ Which method do you prefer? Why?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/elementary-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Elementary Algebra 2e
• Publication date: Apr 22, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/elementary-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/elementary-algebra-2e/pages/10-3-solve-quadratic-equations-using-the-quadratic-formula

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• The Quadratic Formula

What it is, what it does, and how to use it

What is the Quadratic Formula?

The quadratic formula is:

What does this formula tell us?

The quadratic formula calculates the solutions of any quadratic equation.

What is a quadratic equation?

A quadratic equation is an equation that can be written as ax ² + bx + c where a ≠ 0 . In other words, a quadratic equation must have a squared term as its highest power.

Examples of quadratic equations

$$ y = 5x^2 + 2x + 5 \\ y = 11x^2 + 22 \\ y = x^2 - 4x +5 \\ y = -x^2 + + 5 $$

Non Examples

$$ y = 11x + 22 \\ y = x^3 -x^2 +5x +5 \\ y = 2x^3 -4x^2 \\ y = -x^4 + 5 $$

Ok, but what is a 'solution'?

Well a solution can be thought in two ways:

Example of the quadratic formula to solve an equation

Use the formula to solve theQuadratic Equation: $$ y = x^2 + 2x + 1 $$ .

Just substitute a,b, and c into the general formula:

$$ a = 1 \\ b = 2 \\ c = 1 $$

Below is a picture representing the graph of y = x² + 2x + 1 and its solution.

Quadratic Formula Song

A catchy way to remember the quadratic formula is this song (pop goes the weasel).

Practice Problems

Use the quadratic formula to find the solutions to the following equation: y = x² − 2x + 1 and its solution.

Use the quadratic formula to find the solutions to the following equation: y = x² − x − 2 and its solution.

In this quadratic equation, y = x² − x − 2 and its solution:

Use the quadratic formula to find the solutions to the following equation: y = x² − 1 and its solution.

In this quadratic equation, y = x² − 1 and its solution:

Calculate the solutions of the the quadratic equation below by using the quadratic formula : y = x² + 2x − 3 and its solution.

In this quadratic equation, y = x² + 2x − 3 and its solution:

Below is a picture of the graph of the quadratic equation and its two solutions.

Use the quadratic formula to find the solutions to the following equation: y = x² + 4x − 5 and its solution.

In this quadratic equation, y = x² + 4x − 5 and its solution:

Use the quadratic formula to find the solutions to the following equation: y = x² − 4x + 5 and its solution.

In this quadratic equation, y = x² − 4x + 5 and its solution:

Below is a picture of this quadratic's graph.

• Solving Quadratic Equations
• Solve Quadratic Equations By Factoring
• Discriminant
• Completing the Square in Math
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• Formula for Sum and Product of Roots
• Quadratic Inequalities

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• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• 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.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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How to Use the Quadratic Formula to Find Solutions to Quadratic Equations

Last Updated: April 28, 2023 References

This article was co-authored by David Jia and by wikiHow staff writer, Amber Crain . 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 7 references cited in this article, which can be found at the bottom of the page. This article has been viewed 39,540 times.

You can use a few different techniques to solve a quadratic equation and the quadratic formula is one of them. The coolest thing about the formula is that it always works. You can apply it to any quadratic equation out there and you'll get an answer every time. That's not the case with the other techniques! The second coolest thing about the quadratic formula: it's easy to use. In this article, we'll walk you through the entire process from start to finish so you can crush your next algebra exam.

See if the equation equals zero.

Convert the equation to standard form.

Identify the coefficients.

Plug the coefficients into the quadratic formula.

Use the order of operations to simplify the formula.

Simplify the radical.

• Group the pairs: (2 x 2) (2 x 2). There are four 2s, so 4 goes outside the radical sign.

Reduce the problem.

• -6 ÷ 2 = -3

• These are your final answers. [9] X Research source

Circle your answer(s).

Memorize the quadratic formula.

• Sing these lyrics to the tune of Pop Goes the Weasel : X is equal to negative B Plus or minus the square root Of B-squared minus four A C All over two A
• If songs aren't your thing, try memorizing this story instead: A negative boy was thinking yes or no about going to a party. At the party, he talked to a square boy but not to the 4 awesome cats. It was all over at 2 am. [10] X Research source

Community Q&A

You Might Also Like

• ↑ https://www.mesacc.edu/~scotz47781/mat120/notes/quad_formula/quad_formula.html
• ↑ https://www.youtube.com/watch?v=i7idZfS8t8w&t=59s
• ↑ https://www.mathsisfun.com/algebra/quadratic-equation.html
• ↑ https://www.youtube.com/watch?v=i7idZfS8t8w&t=129s
• ↑ https://www.mesacc.edu/~scotz47781/mat120/notes/radicals/simplify/simplifying.html
• ↑ https://www.mesacc.edu/~scotz47781/mat120/notes/radicals/simplify/images/examples/prime_factorization.html
• ↑ https://www.chem.tamu.edu/class/fyp/mathrev/mr-quadr.html

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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:

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

How To Solve Quadratic Equations By Factoring - Quick & Simple! | Algebra Online Course

Quadratic equations have a rich history dating back to ancient Babylon. The difference of squares method is an effective solution for certain equations. Other techniques include the AC method and basic factoring . These equations have practical applications in physics, economics, and engineering.

People in old Babylon figured out how to solve complex math problems over 3,900 years ago . Students still learn these important math skills in algebra classes today. One special trick for solving certain math problems is called the difference of squares . We'll look at these math problems and the difference of squares method, including where it came from, why it's useful, and how to do it.

The story of these math problems goes back a very long time . From ancient times to now, these equations have interested many people and helped math grow. This manual explains quadratic equations clearly for both algebra students and math enthusiasts.

The History of Second-Degree Equations

Quadratic equations have been in use for hundreds of years . Many different people and cultures have worked with these equations over time.

Ancient Beginnings and Babylonian Math

Quadratic equations first appeared in ancient Middle Eastern cultures . By 1900 BC, people in Babylon were already able to tackle quadratic challenges. Ancient methods like completing the square were employed, as evidenced by ancient clay tablets such as YBC 6967 .

The Babylonians used a number system based on 60 , which seems odd to us now. But this system let them do hard math very well. They often used quadratic equations for real problems, like figuring out areas and lengths for building things.

Islamic Golden Age and Al-Khwarizmi's Work

Persian mathematician Muhammad ibn Musa al-Khwarizmi made significant progress in algebra during the 8th century CE. A structured approach to solving quadratic equations was introduced in Al-Khwarizmi's text "Al-Jabr wa'l-muqabala" .

Al-Khwarizmi created a math technique called "completing the square," which teachers still use in modern classrooms. In his work, Al-Khwarizmi categorized quadratic equations into three groups and explained solutions for each. His work helped make algebra its own part of math.

Learning about the Difference of Squares Approach

For certain quadratic equations, the difference of squares method offers an effective solution . What does this method involve, and how can we apply it?

How do we define the Squares' Difference?

A specific type of quadratic equation is known as the squares' difference. Its typical form is a² - b² , with a and b generally representing variables or numerical values. We can always break this down into (a + b)(a - b) .

To illustrate, x² - 25 = 0 fits the squares' difference pattern as it's equivalent to x² - 5² = 0. We can break it down into (x + 5)(x - 5) = 0 by applying the square difference rule.

Applying the Square Difference Technique

To apply the square difference technique:

• See if the equation matches the pattern a² - b² = 0.
• When it does, split it into (a + b)(a - b) = 0.
• Make each section zero and find the value of the unknown.

We'll work out x² - 9 = 0. This equation looks like a square minus another square (x² - 3²). Breaking it down, we get (x + 3)(x - 3) = 0 . When we make each part zero, we find x = -3 or x = 3.

Improving Your Skills in Solving Second-Degree Equations

Subtracting squares is only one technique for tackling quadratic problems. We'll explore additional strategies and their ideal applications.

The AC Method: A Useful Tool

The AC strategy, also known as "coefficient splitting," is used to break down quadratic expressions with three terms , written as ax² + bx + c. It's especially helpful when a is not 1.

Follow these steps to use the AC method:

• Start by multiplying a and c to get what we call the "target number."
• Look for two numbers that multiply to give the last term and add up to the middle term's coefficient.
• Substitute the central expression in the formula using these two figures.
• Arrange the terms into groups and remove shared elements.

When factoring 2x² + 7x + 3, we discover that 1 and 6 produce 6 (2 × 3) when multiplied and sum to 7. The equation in factored form becomes (2x + 1)(x + 3) = 0 .

Picking the Right Factoring Approach

Select the most suitable factoring strategy depending on the quadratic equation's appearance.

• Square Difference Method : Apply when solving equations in the form a² - b² = 0.
• AC Method : Use for tricky quadratic expressions, especially those where x² isn't 1.
• Basic Factoring : Start with this for simple quadratic equations where x² has a coefficient of 1.
• Quadratic Formula : Resort to this when other methods fail, as it works for all quadratic equations.

Remember, practice helps. Solve lots of equations to get better at picking the right method.

Practical Applications of Quadratic Equations

Equations with squared terms are useful in real-world situations , not just math class. These concepts have practical applications across various disciplines and enhance our mathematical understanding.

Real-world Uses of Quadratic Formulas

Second-degree equations appear frequently in everyday situations. They're used in:

• Physics : To study how things move when thrown
• Economics : To look at supply and demand
• Engineering : To design curved structures

Businesses often rely on quadratic functions to maximize profits or minimize costs , finding optimal solutions. These mathematical results guide important corporate decisions.

Upcoming Trends in Quadratic Problem-Solving

Learning about quadratic formulas remains valuable in school and various fields. New tools, like online factoring calculators , make it easier to solve hard equations.

But understanding the basic ideas is still important. As computers and AI get better, being able to think about math ideas becomes even more valuable. The next big math discovery might come from a new way of looking at quadratic equations.

Conclusion: What You've Learned About Second-Degree Equations

The way we handle squared equations has evolved greatly since ancient times. From Al-Khwarizmi's organized approach to modern factoring methods, these equations keep challenging and inspiring math students and experts.

In math studies, remember that grasping the concept of squared equations is more important than just knowing the formulas. The goal is to improve your problem-solving skills and deepen your grasp of mathematical concepts. Feel confident when you see a quadratic equation. Whether you use factoring, square completion, or another method, you're part of a math tradition that's thousands of years old.

Try to solve a quadratic equation using a method you haven't used before. Experimenting with different approaches can lead to unexpected insights and learning opportunities.

In math, like in life, sometimes how you get there is just as important as where you end up.

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