ORCCA The Quadratic Formula

Section 7.2 The Quadratic Formula

Objectives: PCC Course Content and Outcome Guide

MTH 65 CCOG 3.b
MTH 65 CCOG 3.g

We have learned how to solve certain quadratic equations using the square root property. In this section, we will learn another method, the quadratic formula.

Figure 7.2.1. Alternative Video Lesson

Subsection 7.2.1 Solving Quadratic Equations with the Quadratic Formula

The standard form for a quadratic equation is

\begin{equation*} ax^2+bx+c=0 \end{equation*}

where $a$ is some nonzero number.

When $b=0$ and the equation's form is $ax^2+c=0\text{,}$ then we can simply use the square root property to solve it. For example, $x^2-4=0$ leads to $x^2=4\text{,}$ which leads to $x=\pm2\text{,}$ a solution set of $\{-2,2\}\text{.}$

But can we solve equations where $b\neq0\text{?}$ A general method for solving a quadratic equation is to use what is known as the quadratic formula.

Fact 7.2.2. The Quadratic Formula.

For any quadratic equation $ax^2+bx+c=0$ where $a\neq0\text{,}$ the solutions are given by

\begin{equation*} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{equation*}

As we have seen from solving quadratic equations, there can be at most two solutions. Both of the solutions are included in the quadratic formula with the $\pm$ symbol. We could write the two solutions separately:

\begin{align*} x=\frac{-b-\sqrt{b^2-4ac}}{2a}\amp\amp\text{ or }\amp\amp x=\frac{-b+\sqrt{b^2-4ac}}{2a} \end{align*}

This method for solving quadratic equations will work to solve every quadratic equation. It is most helpful when $b\ne0$.

Example 7.2.3.

Linh is in a physics class that launches a tennis ball from a rooftop that is $90.2$ feet above the ground. They fire it directly upward at a speed of $14.4$ feet per second and measure the time it takes for the ball to hit the ground below. We can model the height of the tennis ball, $h\text{,}$ in feet, with the quadratic equation $h=-16x^2+14.4x+90.2\text{,}$ where $x$ represents the time in seconds after the launch. According to the model, when should the ball hit the ground? Round the time to one decimal place.

The ground has a height of $0$ feet. Substituting $0$ for $h$ in the equation, we have this quadratic equation:

\begin{equation*} 0=-16x^2+14.4x+90.2 \end{equation*}

We cannot solve this equation with the square root property, so we will use the quadratic formula. First we will identify that $\highlight{a=-16}\text{,}$ $\highlight{b=14.4}$ and $\highlight{c=90.2}\text{,}$ and substitute them into the formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{14.4})\pm\sqrt{(\substitute{14.4})^2-4(\substitute{-16})(\substitute{90.2})}}{2(\substitute{-16})}\\ x\amp=\frac{-14.4\pm\sqrt{207.36-(-5772.8)}}{-32}\\ x\amp=\frac{-14.4\pm\sqrt{207.36+5772.8}}{-32}\\ x\amp=\frac{-14.4\pm\sqrt{5980.16}}{-32}\\ \end{align*}

These are the exact solutions but because we have a context we want to approximate the solutions with decimals.

\begin{align*} x\amp\approx-1.966\text{ or }x\approx2.866 \end{align*}

We don't use the negative solution because a negative time does not make sense in this context. The ball will hit the ground approximately $2.9$ seconds after it is launched.

The quadratic formula can be used to solve any quadratic equation, but it requires that you don't make any slip-up with remembering the formula, that you correctly identify $a\text{,}$ $b\text{,}$ and $c\text{,}$ and that you don't make any arithmetic mistakes when you calculate and simplify. We recommend that you always check whether you can use the square root property before using the quadratic formula. Here is another example.

Example 7.2.4.

Solve for $x$ in $2x^2-9x+5=0\text{.}$

Explanation

First, we check and see that we cannot use the square root property (because $b\neq0$) so we will use the quadratic formula. Next we identify that $\highlight{a=2}\text{,}$ $\highlight{b=-9}$ and $\highlight{c=5}\text{.}$ We substitute them into the quadratic formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{-9})\pm\sqrt{(\substitute{-9})^2-4(\substitute{2})(\substitute{5})}}{2(\substitute{2})}\\ x\amp=\frac{9\pm\sqrt{81-40}}{4}\\ x\amp=\frac{9\pm\sqrt{41}}{4} \end{align*}

This is fully simplified because we cannot simplify $\sqrt{41}$ or reduce the fraction. The solution set is $\left\{\frac{9-\sqrt{41}}{4},\frac{9+\sqrt{41}}{4}\right\}\text{.}$ We do not have a context here so we leave the solutions in their exact form.

When a quadratic equation is not in standard form we must convert it before we can identify the values of $a\text{,}$ $b$ and $c\text{.}$ We will show that in the next example.

Example 7.2.5.

Solve for $x$ in $x^2=-10x-3\text{.}$

Explanation

First, we convert the equation into standard form by adding $10x$ and $3$ to each side of the equation:

\begin{equation*} x^2+10x+3=0 \end{equation*}

Next, we check that we cannot use the square root property so we will use the quadratic formula. We identify that $\highlight{a=1}\text{,}$ $\highlight{b=10}$ and $\highlight{c=3}\text{.}$ We substitute them into the quadratic formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{10}\pm\sqrt{(\substitute{10})^2-4(\substitute{1})(\substitute{3})}}{2(\substitute{1})}\\ x\amp=\frac{-10\pm\sqrt{100-12}}{2}\\ x\amp=\frac{-10\pm\sqrt{88}}{2} \end{align*}

We notice that the radical can be simplified:

\begin{align*} x\amp=\frac{-10\pm2\sqrt{22}}{2}\\ x\amp=\frac{-10}{2}\pm\frac{2\sqrt{22}}{2}\\ x\amp=-5\pm\sqrt{22} \end{align*}

The solution set is $\{-5-\sqrt{22}, -5+\sqrt{22}\}\text{.}$

Remark 7.2.6.

The irrational solutions to quadratic equations can be checked, although doing so can sometimes involve a lot of simplification and is not shown throughout this section. As an example, to check the solution of $-5+\sqrt{22}$ from Example 7.2.5, we would replace $x$ with $-5+\sqrt{22}$ and check that the two sides of the equation are equal. This check is shown here:

\begin{align*} x^2\amp=-10x-3\\ (\substitute{-5+\sqrt{22}})^2\amp\stackrel{?}{=}-10(\substitute{-5+\sqrt{22}})-3\\ (-5)^2+2(-5)(\sqrt{22})+(\sqrt{22})^2\amp\stackrel{?}{=}-10(-5+\sqrt{22})-3\\ 25-10\sqrt{22}+22 \amp\stackrel{?}{=}50-10\sqrt{22}-3\\ 47-10\sqrt{22}\amp\stackrel{\checkmark}{=}47-10\sqrt{22} \end{align*}

When the radicand from the quadratic formula, $b^2-4ac\text{,}$ which is called the discriminant, is a negative number, the quadratic equation has no real solution. Example 7.2.7 shows what happens in this case.

Example 7.2.7.

Solve for $y$ in $y^2-4y+8=0\text{.}$

Explanation

Identify that $\highlight{a=1}\text{,}$ $\highlight{b=-4}$ and $\highlight{c=8}\text{.}$ We will substitute them into the quadratic formula:

\begin{align*} y\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-(\substitute{-4})\pm\sqrt{(\substitute{-4})^2-4(\substitute{1})(\substitute{8})}}{2(\substitute{1})}\\ \amp=\frac{4\pm\sqrt{16-32}}{2}\\ \amp=\frac{4\pm\sqrt{-16}}{2} \end{align*}

The square root of a negative number is not a real number, so we will simply state that this equation has no real solutions.

Sometimes a radical equation gives rise to a quadratic equation, and the quadratic formula is useful.

Example 7.2.8.

Solve for $z$ in $\sqrt{z}+2=z\text{.}$

Explanation

We will isolate the radical first, and then square both sides.

\begin{align*} \sqrt{z}+2\amp=z\\ \sqrt{z}\amp=z-2\\ \left(\sqrt{z}\right)^{\highlight{2}}\amp=(z-2)^{\highlight{2}}\\ z\amp=z^2-4z+4\\ 0\amp=z^2-5z+4\\ z\amp=\frac{5\pm\sqrt{(-5)^2-4(1)(4)}}{2}\\ \amp=\frac{5\pm\sqrt{25-16}}{2}\\ \amp=\frac{5\pm\sqrt{9}}{2}\\ \amp=\frac{5\pm3}{2} \end{align*}

\begin{align*} z\amp=\frac{5-3}{2}\amp\text{ or }\amp\amp z\amp=\frac{5+3}{2}\\ z\amp=1\amp\text{ or }\amp\amp z\amp=4 \end{align*}

Because we squared both sides of an equation, we must check both solutions.

\begin{align*} \sqrt{\substitute{1}}+2\amp\stackrel{?}{=}1\amp\sqrt{\substitute{4}}+2\amp\stackrel{?}{=}4\\ 1+2\amp\stackrel{\text{no}}{=}1\amp2+2\amp\stackrel{\checkmark}{=}4 \end{align*}

It turned out that $1$ is an extraneous solution, but $4$ is a valid solution. So the equation has one solution: $4\text{.}$ The solution set is $\{4\}\text{.}$

Example 7.2.9.

Solve the equation $\sqrt{2n-6}=1+\sqrt{n-2}$ for $n\text{.}$

Explanation

We cannot isolate two radicals, so we will simply square both sides, and later try to isolate the remaining radical.

\begin{align*} \sqrt{2n-6}\amp=1+\sqrt{n-2}\\ \left(\sqrt{2n-6}\right)^{\highlight{2}}\amp=\left(1+\sqrt{n-2}\right)^{\highlight{2}}\\ 2n-6\amp=1^2+2\sqrt{n-2}+\left(\sqrt{n-2}\right)^2\\ 2n-6\amp=1+2\sqrt{n-2}+n-2\\ 2n-6\amp=2\sqrt{n-2}+n-1\\ n-5\amp=2\sqrt{n-2}\\ \end{align*}

Note here that we can leave the factor of $2$ next to the radical. We will square the $2$ also.

\begin{align*} (n-5)^{\highlight{2}}\amp=\left(2\sqrt{n-2}\right)^{\highlight{2}}\\ n^2-10n+25\amp=4(n-2)\\ n^2-10n+25\amp=4n-8\\ n^2-14n+33\amp=0 \end{align*}

\begin{align*} n\amp=\frac{14\pm\sqrt{14^2-4(1)(33)}}{2}\\ \amp=\frac{14\pm\sqrt{196-132}}{2}\\ \amp=\frac{14\pm\sqrt{64}}{2}\\ \amp=\frac{14\pm8}{2} \end{align*}

\begin{align*} n\amp=\frac{14-8}{2}\amp\text{ or }\amp\amp n\amp=\frac{14+8}{2}\\ n\amp=3\amp\text{ or }\amp\amp n\amp=11 \end{align*}

So our two potential solutions are $3$ and $11\text{.}$ We should now verify that they truly are solutions.

\begin{align*} \sqrt{2(\substitute{3})-6}\amp\stackrel{?}{=}1+\sqrt{\substitute{3}-2}\amp\sqrt{2(\substitute{11})-6}\amp\stackrel{?}{=}1+\sqrt{\substitute{11}-2}\\ \sqrt{6-6}\amp\stackrel{?}{=}1+\sqrt{1}\amp\sqrt{22-6}\amp\stackrel{?}{=}1+\sqrt{9}\\ \sqrt{0}\amp\stackrel{?}{=}1+1\amp\sqrt{16}\amp\stackrel{?}{=}1+3\\ 0\amp\stackrel{\text{no}}{=}2\amp4\amp\stackrel{\checkmark}{=}4 \end{align*}

So, $11$ is the only solution. The solution set is $\{11\}\text{.}$

Reading Questions 7.2.2 Reading Questions

1.

What is the formula for the discriminant? (The part of the quadratic formula inside the radical.)

2.

Are there any kinds of quadratic equations where the quadratic formula is not the best tool to use?

3.

Given a quadratic euqation, will the quadratic formula always show you two real solutions?

Exercises 7.2.3 Exercises

Review and Warmup

1.

Evaluate $\displaystyle{{\frac{-5A+5B+7}{6A-9B}}}$ for $A=2$ and $B=-4\text{.}$

2.

Evaluate $\displaystyle{{\frac{-6C-5c+9}{-6C-3c}}}$ for $C=-10$ and $c=-9\text{.}$

3.

Evaluate the expression $\displaystyle \frac{1}{3} \big( x + 4 \big)^2 - 2$ when $x = -7\text{.}$

4.

Evaluate the expression $\displaystyle \frac{1}{3} \big( x + 4 \big)^2 - 7$ when $x = -7\text{.}$

5.

Evaluate the expression $-16t^{2}+64t+128$ when $t=3\text{.}$

6.

Evaluate the expression $-16t^{2}+64t+128$ when $t=-5\text{.}$

7.

Evaluate the expression ${x^{2}}\text{:}$

For $x=7\text{.}$
For $x=-2\text{.}$

8.

Evaluate the expression ${y^{2}}\text{:}$

For $y=4\text{.}$
For $y=-6\text{.}$

9.

Evaluate each algebraic expression for the given value(s):

$\displaystyle\frac{\sqrt{x}}{y}-\frac{y}{x}\text{,}$ for $x = 25$ and $y = 10\text{:}$

10.

Evaluate each algebraic expression for the given value(s):

$\displaystyle\frac{y}{4 x} - \frac{\sqrt{x}}{3 y}\text{,}$ for $x = 25$ and $y = -4\text{:}$

Solve Quadratic Equations Using the Quadratic Formula

11.

Solve the equation.

${x^{2}+7x+1}= 0$

12.

Solve the equation.

${x^{2}+8x+11}= 0$

13.

Solve the equation.

${20x^{2}+56x+15}=0$

14.

Solve the equation.

${10x^{2}+39x+35}=0$

15.

Solve the equation.

${x^{2}}= {x+1}$

16.

Solve the equation.

${x^{2}}= {5x-5}$

17.

Solve the equation.

${x^{2}+3x-9}= 0$

18.

Solve the equation.

${x^{2}-9x+9}= 0$

19.

Solve the equation.

${2x^{2}+3x-1}= 0$

20.

Solve the equation.

${3x^{2}-x-1}= 0$

21.

Solve the equation.

${4x^{2}-10x-5}= 0$

22.

Solve the equation.

${7x^{2}-2x-1}= 0$

23.

Solve the equation.

${5x^{2}-9x+6}= 0$

24.

Solve the equation.

${3x^{2}+3x+3}= 0$

Solve Quadratic Equations Using an Appropriate Method

25.

Solve the equation.

$3x^2 - 27 = 0$

26.

Solve the equation.

$4x^2 - 16 = 0$

27.

Solve the equation.

${25x^{2}-81}=0$

28.

Solve the equation.

${36x^{2}-25}=0$

29.

Solve the equation.

$4-7r^2 = 1$

30.

Solve the equation.

$0-3r^2 = -7$

31.

Solve the equation.

${x^{2}+5x} = {24}$

32.

Solve the equation.

${x^{2}+4x} = {60}$

33.

Solve the equation.

$\left(x - 9\right)^2 = 64$

34.

Solve the equation.

$\left(x - 7\right)^2 = 16$

35.

Solve the equation.

${x^{2}}= {-9x-16}$

36.

Solve the equation.

${x^{2}}= {-3x+2}$

37.

Solve the equation.

${3x^{2}}= {x+1}$

38.

Solve the equation.

${2x^{2}}= {-\left(5x+1\right)}$

39.

Solve the equation.

$22 - 4 ( r+5 )^2 = 6$

40.

Solve the equation.

$23 - 2 ( t - 8 )^2 = 5$

Radical Equations That Give Rise to Quadratic Equations

41.

Solve the equation.

$\displaystyle{ {\sqrt{t+72}} = {t} }$

42.

Solve the equation.

$\displaystyle{ {\sqrt{2x+15}} = {x} }$

43.

Solve the equation.

$\displaystyle{ {\sqrt{x}+2} = {x} }$

44.

Solve the equation.

$\displaystyle{ {\sqrt{y}+56} = {y} }$

45.

Solve the equation.

$\displaystyle{ {y} = {\sqrt{y+9}+3} }$

46.

Solve the equation.

$\displaystyle{ {r} = {\sqrt{r+1}+5} }$

47.

Solve the equation.

$\displaystyle{ {\sqrt{r}+90} = {r} }$

48.

Solve the equation.

$\displaystyle{ {\sqrt{r}+42} = {r} }$

49.

Solve the equation.

$\displaystyle{ {t} = {\sqrt{t+3}+9} }$

50.

Solve the equation.

$\displaystyle{ {t} = {\sqrt{t+1}+89} }$

51.

Solve the equation.

$\displaystyle{ {\sqrt{51-x}} = {x+5} }$

52.

Solve the equation.

$\displaystyle{ {\sqrt{148-x}} = {x+8} }$

Quadratic Formula Applications

53.

Two numbers’ sum is $-1\text{,}$ and their product is $-42\text{.}$ Find these two numbers.

These two numbers are .

54.

Two numbers’ sum is $-13\text{,}$ and their product is $42\text{.}$ Find these two numbers.

These two numbers are .

55.

Two numbers’ sum is ${7.7}\text{,}$ and their product is ${-25.5}\text{.}$ Find these two numbers.

These two numbers are . (Use a comma to separate your numbers.)

56.

Two numbers’ sum is ${4.7}\text{,}$ and their product is ${3.96}\text{.}$ Find these two numbers.

These two numbers are . (Use a comma to separate your numbers.)

57.

A rectangle’s base is ${6\ {\rm cm}}$ longer than its height. The rectangle’s area is ${112\ {\rm cm^{2}}}\text{.}$ Find this rectangle’s dimensions.

The rectangle’s height is .

The rectangle’s base is .

58.

A rectangle’s base is ${9\ {\rm cm}}$ longer than its height. The rectangle’s area is ${162\ {\rm cm^{2}}}\text{.}$ Find this rectangle’s dimensions.

The rectangle’s height is .

The rectangle’s base is .

59.

A rectangle’s base is ${3\ {\rm in}}$ shorter than four times its height. The rectangle’s area is ${85\ {\rm in^{2}}}\text{.}$ Find this rectangle’s dimensions.

The rectangle’s height is .

The rectangle’s base is .

60.

A rectangle’s base is ${1\ {\rm in}}$ shorter than twice its height. The rectangle’s area is ${15\ {\rm in^{2}}}\text{.}$ Find this rectangle’s dimensions.

The rectangle’s height is .

The rectangle’s base is .

61.

You will build a rectangular sheep pen next to a river. There is no need to build a fence along the river, so you only need to build three sides.

You have a total of $510$ feet of fence to use, and the area of the pen must be $31900$ square feet. Find the dimensions of the pen.

There should be two solutions:

When the width is feet, the length is feet.

62.

You will build a rectangular sheep pen next to a river. There is no need to build a fence along the river, so you only need to build three sides.

You have a total of $470$ feet of fence to use, and the area of the pen must be $27500$ square feet. Find the dimensions of the pen.

There should be two solutions:

When the width is feet, the length is feet.

63.

There is a rectangular lot in the garden, with ${8\ {\rm ft}}$ in length and ${4\ {\rm ft}}$ in width. You plan to expand the lot by an equal length around its four sides, and make the area of the expanded rectangle ${140\ {\rm ft^{2}}}\text{.}$ How long should you expand the original lot in four directions?

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You should expand the original lot by in four directions.

64.

There is a rectangular lot in the garden, with ${9\ {\rm ft}}$ in length and ${7\ {\rm ft}}$ in width. You plan to expand the lot by an equal length around its four sides, and make the area of the expanded rectangle ${195\ {\rm ft^{2}}}\text{.}$ How long should you expand the original lot in four directions?

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You should expand the original lot by in four directions.

65.

One car started at Town A, and traveled due north at $60$ miles per hour. $2$ hours later, another car started at the same spot and traveled due east at $55$ miles per hour. Assume both cars don’t stop, after how many hours since the second car starts would the distance between them be $338$ miles? Round your answer to two decimal places if needed.

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Approximately hours since the second car starts, the distance between those two cars would be $338$ miles.

66.

One car started at Town A, and traveled due north at $65$ miles per hour. $3$ hours later, another car started at the same spot and traveled due east at $40$ miles per hour. Assume both cars don’t stop, after how many hours since the second car starts would the distance between them be $358$ miles? Round your answer to two decimal places if needed.

eps pdf png svg tex

Approximately hours since the second car starts, the distance between those two cars would be $358$ miles.

67.

An object is launched upward at the height of $370$ meters. Its height can be modeled by

\begin{equation*} h=-4.9t^2+90t+370\text{,} \end{equation*}

where $h$ stands for the object’s height in meters, and $t$ stands for time passed in seconds since its launch. The object’s height will be $380$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $380$ meters. Round your answers to two decimal places if needed.

The object’s height would be $380$ meters the first time at seconds, and then the second time at seconds.

68.

An object is launched upward at the height of $400$ meters. Its height can be modeled by

\begin{equation*} h=-4.9t^2+70t+400\text{,} \end{equation*}

where $h$ stands for the object’s height in meters, and $t$ stands for time passed in seconds since its launch. The object’s height will be $430$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $430$ meters. Round your answers to two decimal places if needed.

The object’s height would be $430$ meters the first time at seconds, and then the second time at seconds.

69.

Currently, an artist can sell $280$ paintings every year at the price of ${\$60.00}$ per painting. Each time he raises the price per painting by ${\$5.00}\text{,}$ he sells $5$ fewer paintings every year.

Assume he will raise the price per painting $x$ times, then he will sell $280-5x$ paintings every year at the price of $60+5x$ dollars. His yearly income can be modeled by the equation:

\begin{equation*} i=(60+5x)(280-5x) \end{equation*}

where $i$ stands for his yearly income in dollars. If the artist wants to earn ${\$22{,}500.00}$ per year from selling paintings, what new price should he set?

To earn ${\$22{,}500.00}$ per year, the artist could sell his paintings at two different prices. The lower price is per painting, and the higher price is per painting.

70.

Currently, an artist can sell $210$ paintings every year at the price of ${\$130.00}$ per painting. Each time he raises the price per painting by ${\$10.00}\text{,}$ he sells $5$ fewer paintings every year.

Assume he will raise the price per painting $x$ times, then he will sell $210-5x$ paintings every year at the price of $130+10x$ dollars. His yearly income can be modeled by the equation:

\begin{equation*} i=(130+10x)(210-5x) \end{equation*}

where $i$ stands for his yearly income in dollars. If the artist wants to earn ${\$33{,}300.00}$ per year from selling paintings, what new price should he set?

To earn ${\$33{,}300.00}$ per year, the artist could sell his paintings at two different prices. The lower price is per painting, and the higher price is per painting.

71.

Solve for $x$ in the equation $mx^{2}+nx+p=0\text{.}$