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## Learning objectives

• Identify and choose methods for solving quadratic equations.
• Solve real-world problems using functions by completing the square.

## Introduction

In this section, you will solve quadratic equations using the Quadratic Formula. Most of you are already familiar with this formula from previous mathematics courses. It is probably the most used method for solving quadratic equations. For a quadratic equation in standard form

$ax^2 + bx + c = 0$

The solutions are found using the following formula.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}$

We will start by explaining where this formula comes from and then show how it is applied. This formula is derived by solving a general quadratic equation using the method of completing the square that you learned in the previous section.

$\text{We start with a general quadratic equation}. & & ax^2 + bx + c & = 0\\\text{Subtract the constant term from both sides}. & & ax^2 + bx & = -c\\\text{Divide by the coefficient of the} \ x^2 \ \text{term}. & & x^2 + \frac{b} {a} x & = -\frac{c} {a}\\\text{Rewrite}. & & x^2 + 2\left (\frac{b} {2a}\right ) x & = - \frac{c} {a}\\\text{Add the constant} \left (\frac{b} {2a}\right )^2 \ \text{to both sides}. & & x^2 + 2 \left (\frac{b} {2a}\right ) x + \left (\frac{b} {2a}\right )^2 & = -\frac{c} {a} + \frac{b^2} {4a^2}\\\text{Factor the perfect square trinomial}. & & \left (x + \frac{b} {2a}\right )^2 & = -\frac{4ac} {4a^2} + \frac{b^2} {4a^2}\\\text{Simplify}. & & \left (x + \frac{b} {2a}\right )^2 & = \frac{b^2 - 4ac} {4a^2}\\\text{Take the square root of both sides}. & & x + \frac{b} {2a} & = \sqrt{\frac{b^2 - 4ac} {4a^2}} \ \text{and} \ x + \frac{b} {2a} = - \sqrt{\frac{b^2 - 4ac} {4a^2}}\\\text{Simplify}. & & x + \frac{b} {2a} & = \sqrt{\frac{b^2 - 4ac} {2a}} \ \text{and} \ x + \frac{b} {2a} = - \sqrt{\frac{b^2 - 4ac} {2a}}\\& & x & = - \frac{b} {2a} + \sqrt{\frac{b^2 - 4ac} {2a}}\\ & & x & = - \frac{b} {2a} - \sqrt{\frac{b^2 - 4ac} {2a}}\\& & x & = \frac{-b + \sqrt{b^2 - 4ac}} {2a} \\& & x & = \frac{- b - \sqrt{b^2 - 4ac}} {2a}$

This can be written more compactly as $x = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}$.

You can see that the familiar formula comes directly from applying the method of completing the square. Applying the method of completing the square to solve quadratic equations can be tedious. The quadratic formula is a more straightforward way of finding the solutions.

Applying the quadratic formula basically amounts to plugging the values of $a, b$ and $c$ into the quadratic formula.

Example 1

a) $2x^2 + 3x + 1 = 0$

b) $x^2 - 6x + 5 = 0$

c) $-4x^2 + x + 1 = 0$

Solution

Start with the quadratic formula and plug in the values of $a, b$ and $c$.

a) $\text{Quadratic formula} & & x & = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}\\\text{Plug in the values} \ a = 2, b = 3, c = 1. & & x & = \frac{-3 \pm \sqrt{(3)^2 - 4(2) (1)}} {2(2)}\\\text{Simplify}. & & x & = \frac{-3 \pm \sqrt{9 - 8}} {4} = \frac{-3 \pm \sqrt{1}} {4}\\\text{Separate the two options}. & & x & = \frac{-3 + 1} {4} \ \text{and} \ x = \frac{-3-1} {4}\\\text{Solve}. & & x & = \frac{-2} {4} = - \frac{1} {2} \ \text{and} \ x = \frac{-4} {4} = -1$

Answer $x = -\frac{1} {2}$ and $x = -1$

b) $\text{Quadratic formula}. & & x & = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}\\\text{Plug in the values} \ a = 1, b = -6, c = 5. & & x & = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1) (5)}} {2(1)}\\\text{Simplify}. & & x & = \frac{6 \pm \sqrt{36 - 20}} {2} = \frac{6 \pm \sqrt{16}} {2}\\\text{Separate the two options}. & & x & = \frac{6 + 4} {2} \ \text{and} \ x = \frac{6 - 4} {2}\\\text{Solve} & & x & = \frac{10} {2} = 5 \ \text{and} \ x = \frac{2} {2} = 1$

Answer $x = 5$ and $x = 1$

c) $\text{Quadratic formula}. & & x & = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}\\\text{Plug in the values} \ a = -4, b = 1, c = 1. & & x & = \frac{-1 \pm \sqrt{(1)^2 - 4(-4)(1)}} {2(-4)}\\\text{Simplify}. & & x & = \frac{-1 \pm \sqrt{1 + 16}} {-8} = \frac{-1 \pm \sqrt{17}} {-8}\\\text{Separate the two options}. & & x & = \frac{-1 + \sqrt{17}} {-8} \ \text{and} \ x = \frac{-1-\sqrt{17}} {-8}\\\text{Solve}. & & x & \approx -.39 \ \text{and} \ x \approx .64$

Answer $x \approx -.39$ and $x \approx .64$

Often when we plug the values of the coefficients into the quadratic formula, we obtain a negative number inside the square root. Since the square root of a negative number does not give real answers, we say that the equation has no real solutions. In more advanced mathematics classes, you will learn how to work with “complex” (or “imaginary”) solutions to quadratic equations.

Example 2

Solve the following quadratic equation using the quadratic formula $x^2 + 2x + 7 = 0$

Solution:

a) $\text{Quadratic formula}. & & x & = \frac{b \pm\sqrt{b^2 - 4ac}} {2a}\\\text{Plug in the values} \ a = 1, b = 2, c = 7. & & x & = \frac{-2 \pm\sqrt{(2)^2 - 4(1)(7)}} {2(1)}\\\text{Simplify}. & & x & = \frac{-2 \pm \sqrt{4 - 28}} {2} = \frac{-2 \pm \sqrt{-24}} {2}$

Answer There are no real solutions.

To apply the quadratic formula, we must make sure that the equation is written in standard form. For some problems, we must rewrite the equation before we apply the quadratic formula.

Example 3

a) $x^2 - 6x = 10$

b) $8x^2 = 5x + 6$

Solution:

a) $\text{Rewrite the equation in standard form}. & & x^2 - 6x - 10 & = 0\\\text{Quadratic formula} & & x & = \frac{-b \pm\sqrt{b^2 - 4ac}} {2a}\\\text{Plug in the values} \ a = 1, b = -6, c = -10. & & x & = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1) (-10)}} {2(1)}\\\text{Simplify}. & & x & = \frac{6 \pm \sqrt{36 + 40}} {2} = \frac{6 \pm \sqrt{76}} {2}\\\text{Separate the two options}. & & x & = \frac{6 + \sqrt{76}} {2} \ \text{and} \ x = \frac{6 - \sqrt{76}} {2}\\\text{Solve}. & & x & \approx 7.36 \ \text{and} \ x \approx -1.36$

Answer $x \approx 7.36$ and $x \approx -1.36$

b) $\text{Rewrite the equation in standard form}. & & 8x^2 + 5x + 6 & = 0\\\text{Quadratic formula} & & x & = \frac{-b \pm\sqrt{b^2 - 4ac}} {2a}\\\text{Plug in the values} \ a = 8, b = 5, c = 6. & & x & = \frac{-5 \pm \sqrt{(5)^2 - 4(8)(6)}} {2(8)}\\\text{Simplify}. & & x & = \frac{-5 \pm \sqrt{25 - 192}} {16} = \frac{-5 \pm \sqrt{-167}} {16}$

. This video is not necessarily different from the examples above, but it does help reinforce the procedure of using the quadratic formula to solve equations.

## Finding the Vertex of a Parabola with the Quadratic Formula

Sometimes you get more information from a formula beyond what you were originally seeking. In this case, the quadratic formula also gives us an easy way to locate the vertex of a parabola.

First, recall that the quadratic formula tells us the roots or solutions of the equation $ax^2 + bx + c = 0$. Those roots are

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.$

We can rewrite the fraction in the quadratic formula as

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}.$

Recall that the roots are symmetric about the vertex. In the form above, we can see that the roots of a quadratic equation are symmetric around the $x-$coordinate $-\frac{b}{2a}$ because they move $\frac{\sqrt{b^2-4ac}}{2a}$ units to the left and right (recall the $\pm$ sign) from the vertical line $x = - \frac{b}{2a}$. The image to the right illustrates this for the equation $x^2-2x - 3 = 0$. The roots, -1 and 3 are both 2 units from the vertical line $x = 1$.

## Identify and Choose Methods for Solving Quadratic Equations.

In mathematics, you will need to solve quadratic equations that describe application problems or that are part of more complicated problems. You learned four ways of solving a quadratic equation.

• Factoring.
• Taking the square root.
• Completing the square.

Usually you will not be told which method to use. You will have to make that decision yourself. However, here are some guidelines to which methods are better in different situations.

Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you can factor because this is the fastest method. Many expressions are not factorable so this method is not used very often in practice.

Taking the square root is best used when there is no $x$ term in the equation.

Completing the square can be used to solve any quadratic equation. This is usually not any better than using the quadratic formula (in terms of difficult computations), however it is a very important method for re-writing a quadratic function in vertex form. It is also be used to re-write the equations of circles, ellipses and hyperbolas in standard form (something you will do in algebra II, trigonometry, physics, calculus, and beyond...).

Quadratic formula is the method that is used most often for solving a quadratic equation. When solving directly by taking square root and factoring does not work, this is the method that most people prefer to use.

If you are using factoring or the quadratic formula make sure that the equation is in standard form.

Example 4

a) $x^2 - 4x - 5 = 0$

b) $x^2 = 8$

c) $-4x^2 + x = 2$

d) $25x^2 - 9 = 0$

e) $3x^2 = 8x$

Solution

a) This expression if easily factorable so we can factor and apply the zero-product property:

$\text{Factor}. & & (x - 5)(x + 1) & = 0\\\text{Apply zero-product property}. & & x - 5 & = 0 \ \text{and} \ x + 1 = 0\\\text{Solve}. & & x & = 5 \ \text{and} \ x = -1$

Answer $x = 5$ and $x = -1$

b) Since the expression is missing the $x$ term we can take the square root:

$\text{Take the square root of both sides}. & & x = \sqrt{8} \ \text{and} \ x = -\sqrt{8}$

Answer $x = 2.83$ and $x = -2.83$

c) Rewrite the equation in standard form.

It is not apparent right away if the expression is factorable, so we will use the quadratic formula.

$\text{Quadratic formula} & & x & = \frac{-b \pm\sqrt{b^2 - 4ac}} {2a}\\\text{Plug in the values} \ a = -4, b = 1, c = -2. & & x & = \frac{-1 \pm \sqrt{1^2 - 4 (-4) (-2)}} {2(-4)}\\\text{Simplify}. & & x & = \frac{-1 \pm \sqrt{1 - 32}} {-8} = \frac{-1 \pm \sqrt{-31}} {-8}$

d) This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in this case.

$\text{Add} \ 9 \ \text{to both sides of the equation}. & & 25x^2 & = 9\\\text{Divide both sides by} \ 25. & & x^2 & = \frac{9} {25}\\\text{Take the square root of both sides}. & & x & = \sqrt{\frac{9} {25}} \ \text{and} \ x = - \sqrt{\frac{9} {25}}\\\text{Simplify}. & & x & = \frac{3} {5} \ \text{and} \ x = - \frac{3} {5}$

Answer $x = \frac{3} {5}$ and $x = - \frac{3} {5}$

e) $\text{Rewrite the equation in standard form} & & 3x^2 - 8x & = 0\\\text{Factor out common} \ x \ \text{term}. & & x (3x - 8) & = 0\\\text{Set both terms to zero}. & & x & = 0 \ \text{and} \ 3x = 8\\\text{Solve}. & & x & = 0 \ \text{and} \ x = \frac{8} {3} = 2.67$

Answer $x = 0$ and $x = 2.67$

## Solve Real-World Problems Using Quadratic Functions by any Method

Here are some application problems that arise from number relationships and geometry applications.

Example 5

The product of two positive consecutive integers is 156. Find the integers.

Solution

For two consecutive integers, one integer is one more than the other one.

Define

Let $x \ =$ the smaller integer

$x + 1 \ =$ the next integer

Translate

The product of the two numbers is 156. We can write the equation:

$x(x + 1) = 156$

Solve

$x^2 + x & = 156\\x^2 + x - 156 & = 0$

Apply the quadratic formula with $a = 1, b = 1, c = -156$

$& x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-156)}} {2(1)}\\& x = \frac{-1 \pm \sqrt{625}} {2} = \frac{-1 \pm 25} {2}\\& x = \frac{-1 + 25} {2} \ \text{and} \ x = \frac{-1-25} {2}\\& x = \frac{24} {2} = 12 \ \text{and} \ x = \frac{-26} {2} = -13$

Since we are looking for positive integers take, $x = 12$

Check $12 \times 13 = 156$. The answer checks out.

Example 6

The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square/meters. Find the dimensions of the pool.

Solution:

Draw a sketch

Define

Let $x \ =$ the width of the pool

$x + 10 \ =$ the length of the pool

Translate

The area of a rectangle is A = length $\times$ width, so

$x(x + 10) = 875$

Solve

$x^2 + 10x & = 875\\x^2 + 10x - 875 & = 0$

Apply the quadratic formula with $a = 1, b = 10$ and $c = -875$

$x & = \frac{-10 \pm \sqrt{(10)^2 - 4(1) (-875)}} {2(1)}\\x & = \frac{-10 \pm \sqrt{100 + 3500}} {2}\\x & = \frac{-10 \pm \sqrt{3600}} {2} = \frac{-10 \pm 60} {2}\\x & = \frac{-10 + 60} {2}\ \text{and} \ x = \frac{-10 - 60} {2}\\x & = \frac{50} {2} = 25 \ \text{and} \ x = \frac{-70} {2} = -35$

Since the dimensions of the pools should be positive, then $x = 25\ meters$.

Answer The pool is $25\ meters \times 35\ meters$.

Check $25 \times 35 = 875 \ m^2$. The answer checks out.

Example 7

Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is $200 \ ft^2$. How much fencing does Suzie need?

Solution

Draw a Sketch

Define

Let $x \ =$ the width of the plot

$2x \ =$ the length of the plot

Translate

Area of a rectangle is A = length $\times$ width, so

$x(2x) = 200$

Solve

$2x^2 = 200$

Solve by taking the square root.

$x^2 & = 100\\x & = \sqrt{100} \ \text{and} \ x = - \sqrt{100}\\x & = 10 \ \text{and} \ x = -10$

We take $x = 10$ since only positive dimensions make sense.

The plot of land is $10\ feet \times 20\ feet$.

To fence the garden the way Suzie wants, we need 2 lengths and 4 widths $= 2(20) + 4(10) = 80\ feet$ of fence.

Answer: The fence is 80 feet.

Check $10 \times 20 = 200 \ ft^2$ and $2(20) + 4(10) = 80\ feet$. The answer checks out.

Example 8

An isosceles triangle is enclosed in a square so that its base coincides with one of the sides of the square and the tip of the triangle touches the opposite side of the square. If the area of the triangle is $20 \ in^2$ what is the area of the square?

Solution:

Draw a sketch.

Define

Let $x \ =$ base of the triangle

$x \ =$ height of the triangle

Translate

Area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$, so

$\frac{1}{2} \cdot x \cdot x = 20$

Solve

$\frac{1} {2} x^2 & = 20$

Solve by taking the square root.

$x^2 & = 40\\x & = \sqrt{40} \ \text{and} \ x = -\sqrt{40}\\x & \approx 6.32 \ \text{and} \ x \approx -6.32$

The side of the square is 6.32 inches.

The area of the square is $(6.32)^2 = 40 \ in^2$, twice as big as the area of the triangle.

Answer: Area of the triangle is $40 \ in^2$

Check: It makes sense that the area of the square will be twice that of the triangle. If you look at the figure you can see that you can fit two triangles inside the square.

## Review Questions

1. $x^2 + 4x - 21 = 0$
2. $x^2 - 6x = 12$
3. $3x^2 - \frac{1} {2} x = \frac{3} {8}$
4. $2x^2 + x - 3 = 0$
5. $-x^2 - 7x + 12 = 0$
6. $-3x^2 + 5x = 0$
7. $4x^2 = 0$
8. $x^2 + 2x + 6 = 0$

1. $x^2 - x = 6$
2. $x^2 - 12 = 0$
3. $-2x^2 + 5x - 3 = 0$
4. $x^2 + 7x - 18 = 0$
5. $3x^2 + 6x = -10$
6. $-4x^2 + 4000x = 0$
7. $-3x^2 + 12x + 1 = 0$
8. $x^2 + 6x + 9 = 0$
9. $81x^2 + 1 = 0$
10. $-4x^2 + 4x = 9$
11. $36x^2 - 21 = 0$
12. $x^2 - 2x - 3 = 0$
13. The product of two consecutive integers is 72. Find the two numbers.
14. The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers.
15. The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches, find its dimensions.
16. Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece of wood is $4\ feet \times 8\ feet$ and the cut off part is $\frac{1}{3}$ of the total area of the plywood sheet. What is the length of the side of the square?
17. Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of the patio is $192 \ ft^2$ and the length is 4 feet longer than the width. Find how much fencing Mike will need.

1. $x = -7, x = 3$
2. $x = -1.58, x = 7.58$
3. $x = -.28, x = .45$
4. $x = -1.5, x = 1$
5. $x = -8.42, x = 1.42$
6. $x = 1, x = \frac{2}{3}$
7. $x = 0, x = \frac{1}{4}$
8. No real solution
9. $x = -2, x = 3$
10. $x = -3.46, x = 3.46$
11. $x = 1, x = 1.5$
12. $x = -9, x = 2$
13. No real solution
14. $x = 0, x = 1000$
15. $x = -.08, x = 4.08$
16. $x = -3$
17. No real solution
18. No real solution
19. $x = -.76, x = .76$
20. $x = -1, x = 3$
21. 8 and 9
22. 5 and 7
23. 7 in and 10 in
24. $\text{side} = 3.27 \ ft$
25. 40 feet of fencing.

Feb 23, 2012

Sep 15, 2014