## Section8.7Solving Quadratic Equations Chapter Review

### Subsection8.7.1Solving Quadratic Equations by Factoring

In SectionÂ 8.1 we covered the zero product property and learned an algorithm for solving quadratic equations by factoring.

###### Example8.7.1Solving Using Factoring

Solve the quadratic equations using factoring.

1. $x^2-2x-15=0$

2. $4x^2-40x=-96$

3. $6x^2+x-12=0$

4. $(x-3)(x+2)=14$

5. $x^3-64x=0$

Explanation
1. Use factor pairs.

\begin{align*} x^2-2x-15\amp=0\\ (x-5)(x+3)\amp=0 \end{align*}
\begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+3\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-3 \end{align*}

So the solution set is $\{5,-3\}\text{.}$

2. Start by putting the equation in standard form and factoring out the greatest common factor.

\begin{align*} 4x^2-40x\amp=-96\\ 4x^2-40x+96\amp=0\\ 4\left(x^2-10x+24\right)\amp=0\\ 4(x-6)(x-4)\amp=0 \end{align*}
\begin{align*} x-6\amp=0 \amp\text{ or }\amp\amp x-4\amp=0\\ x\amp=6 \amp\text{ or }\amp\amp x\amp=4 \end{align*}

So the solution set is $\{4,6\}\text{.}$

3. Use the AC method.

\begin{align*} 6x^2+x-12\amp=0\\ \end{align*}

Note that $a\cdot c=-72$ and that $\highlight{9\cdot-8}=-72$ and $\highlight{9-8}=1$

\begin{align*} 6x^2\substitute{+9x-8x}-12\amp=0\\ \left(6x^2+9x\right)+\left(-8x-12\right)\amp=0\\ \highlight{3x}\left(2x+3\right)\mathbin{\highlight{-4}}\left(2x+3\right)\amp=0\\ \left(2x+3\right)\highlight{\left(3x-4\right)}\amp=0 \end{align*}
\begin{align*} 2x+3\amp=0 \amp\text{ or }\amp\amp \highlight{3x-4}\amp=0\\ x\amp=-\frac{3}{2} \amp\text{ or }\amp\amp x\amp=\highlight{\frac{4}{3}} \end{align*}

So the solution set is $\left\{-\frac{3}{2},\frac{4}{3}\right\}\text{.}$

4. Start by putting the equation in standard form.

\begin{align*} (x-3)(x+2)\amp=14\\ x^2-x-6\amp=14\\ x^2-x-20\amp=0\\ (x-5)(x+4)\amp=0 \end{align*}
\begin{align*} x-5\amp=0 \amp\text{ or }\amp\amp x+4\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=-4 \end{align*}

So the solution set is $\{5,-4\}\text{.}$

5. Even though this equation has a power higher than $2\text{,}$ we can still find all of its solutions by following the algorithm. Start by factoring out the greatest common factor.

\begin{align*} x^3-64x\amp=0\\ x\left(x^2-64\right)\amp=0\\ x(x-8)(x+8)\amp=0 \end{align*}
\begin{align*} x\amp=0 \amp\text{ or }\amp\amp x-8\amp=0 \amp\text{ or }\amp\amp x+8\amp=0\\ x\amp=0 \amp\text{ or }\amp\amp x\amp=8 \amp\text{ or }\amp\amp x\amp=-8 \end{align*}

So the solution set is $\{0,8,-8\}\text{.}$

### Subsection8.7.2Square Root Properties

In SectionÂ 8.2 we covered the definition of a square root, how to estimate and simplify square roots, multiplication and division properties of square roots, and rationalizing the denominator.

###### Example8.7.2Estimating Square Roots

Estimate the value of $\sqrt{28}$ without a calculator.

Explanation

To estimate $\sqrt{28}\text{,}$ we can find the nearest perfect squares that are whole numbers on either side of $28\text{.}$ Recall that the perfect squares are $1, 4, 9, 16, 25, 36, 49, 64,\dots$ The perfect square that is just below $28$ is $25$ and the perfect square just above $28$ is $36\text{.}$ This tells us that $\sqrt{28}$ is between $\sqrt{25}$ and $\sqrt{36}\text{,}$ or between $5$ and $6\text{.}$ We can also say that $\sqrt{28}$ is closer to $5$ than $6$ because $28$ is closer to $25\text{,}$ so we think $5.2$ or $5.3$ would be a good estimate.

On the calculator we can see that $\sqrt{28}\approx5.29\text{,}$ so our guess was very close to reality.

###### Example8.7.3Multiplication and Division Properties of Square Roots

Simplify the expressions using the multiplication and division properties of square roots.

1. $\sqrt{18}\cdot\sqrt{2}\text{.}$

2. $\frac{\sqrt{18}}{\sqrt{2}}\text{.}$

Explanation
1. \begin{aligned}[t] \sqrt{18}\cdot\sqrt{2}\amp=\sqrt{18\cdot2}\\ \amp=\sqrt{36}\\ \amp=6\end{aligned}

2. \begin{aligned}[t] \frac{\sqrt{18}}{\sqrt{2}}\amp=\sqrt{\frac{18}{2}}\\ \amp=\sqrt{9}\\ \amp=3\end{aligned}

###### Example8.7.4Simplifying Square Roots

Simplify the expression $\sqrt{54}\text{.}$

Explanation

Recall that the perfect squares are $1, 4, 9, 16, 25, 36, 49, 64,\dots$ To simplify the $\sqrt{54}\text{,}$ we need to look at that list and find the largest perfect square the goes into $54$ evenly. In this case, it is $\highlight{9}\text{.}$ We then break up $54$ into two factors $\highlight{9}$ and $6\text{,}$ and we have:

\begin{align*} \sqrt{54}\amp=\sqrt{\highlight{9}\cdot6}\\ \amp=\sqrt{\highlight{9}}\cdot\sqrt{6}\\ \amp=\highlight{3}\sqrt{6} \end{align*}

Since $6$ has no perfect square factors, we can stop.

###### Example8.7.5Multiplying Square Root Expressions

Simplify the expression $\sqrt{50}\cdot\sqrt{27}\text{.}$

Explanation

Note that $25$ is a perfect-square factor of $50\text{,}$ and that $9$ is a perfect-square factor of $27\text{.}$ Now we have:

\begin{align*} \sqrt{50}\cdot\sqrt{27}\amp=\sqrt{\highlight{25}\cdot2}\cdot\sqrt{\lighthigh{9}\cdot3}\\ \amp=\sqrt{\highlight{25}}\cdot\sqrt{2}\cdot\sqrt{\lighthigh{9}}\cdot\sqrt{3}\\ \amp=\highlight{5}\cdot\sqrt{2}\cdot\lighthigh{3}\cdot\sqrt{3}\\ \amp=15\sqrt{6} \end{align*}
###### Example8.7.6Adding and Subtracting Square Root Expressions

Simplify the expression $\sqrt{32}+\sqrt{50}\text{.}$

Explanation

Recall that radicals can only be added if the radicands match identically, so we cannot initially combine these two terms. However, if we simplify first, we may be able to add terms later. Note that $16$ is a perfect-square factor of $32\text{,}$ and that $25$ is a perfect-square factor of $50\text{.}$

\begin{align*} \sqrt{32}+\sqrt{50}\amp=\sqrt{\highlight{16}\cdot2}+\sqrt{\lighthigh{25}\cdot2}\\ \amp=\sqrt{\highlight{16}}\cdot\sqrt{2}+\sqrt{\lighthigh{25}}\cdot\sqrt{2}\\ \amp=\highlight{4}\sqrt{2}+\lighthigh{5}\sqrt{2}\\ \amp=\highlight{9}\sqrt{2} \end{align*}
###### Example8.7.7Rationalizing the Denominator

Rationalize the denominator in the expression $\frac{2}{\sqrt{6}}\text{.}$

Explanation
\begin{align*} \frac{2}{\sqrt{6}}\amp=\frac{2}{\sqrt{6}}\multiplyright{\frac{\sqrt{6}}{\sqrt{6}}}\\ \amp=\frac{2\multiplyright{\sqrt{6}}}{\sqrt{6}\multiplyright{\sqrt{6}}}\\ \amp=\frac{2\sqrt{6}}{6}\\ \amp=\frac{\sqrt{6}}{3} \end{align*}
###### Example8.7.8More Complicated Square Roots

Expand $\left(\sqrt{5}+\sqrt{3}\right)^2\text{.}$

Explanation

We will use the FOIL method to expand this expression:

\begin{align*} \left(\sqrt{5}+\sqrt{3}\right)^2\amp=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\\ \amp=\left(\sqrt{5}\right)^2+\sqrt{5}\sqrt{3}+\sqrt{3}\sqrt{5}+\left(\sqrt{3}\right)^2\\ \amp=5+\sqrt{15}+\sqrt{15}+3\\ \amp=8+2\sqrt{15} \end{align*}

### Subsection8.7.3Solving Quadratic Equations by Using a Square Root

In SectionÂ 8.3 we covered how to solve quadratic equations using the square root property and how to use the Pythagorean Theorem.

###### Example8.7.9Solving Quadratic Equations Using the Square Root Property

Solve for $w$ in $3(2-w)^2-24=0\text{.}$

Explanation

It's important here to suppress any urge you may have to expand the squared binomial. We begin by isolating the squared expression.

\begin{align*} 3(2-w)^2-24\amp=0\\ 3(2-w)^2\amp=24\\ (2-w)^2\amp=8 \end{align*}

Now that we have the squared expression isolated, we can use the square root property.

\begin{align*} 2-w\amp=-\sqrt{8}\amp\text{or}\amp\amp2-w\amp=\sqrt{8}\\ 2-w\amp=-\sqrt{\highlight{4}\cdot2}\amp\text{or}\amp\amp2-w\amp=\sqrt{\highlight{4}\cdot2}\\ 2-w\amp=-\sqrt{\highlight{4}}\cdot\sqrt{2}\amp\text{or}\amp\amp2-w\amp=\sqrt{\highlight{4}}\cdot\sqrt{2}\\ 2-w\amp=-\highlight{2}\sqrt{2}\amp\text{or}\amp\amp2-w\amp=\highlight{2}\sqrt{2}\\ -w\amp=-2\sqrt{2}-2\amp\text{or}\amp\amp-w\amp=2\sqrt{2}-2\\ w\amp=2\sqrt{2}+2\amp\text{or}\amp\amp w\amp=-2\sqrt{2}+2 \end{align*}

The solution set is $\left\{2\sqrt{2}+2,-2\sqrt{2}+2\right\}\text{.}$

###### Example8.7.10The Pythagorean Theorem

Faven was doing some wood working in her garage. She needed to cut a triangular piece of wood for her project that had a hypotenuse of $16$ inches, and the sides of the triangle should be equal in length. How long should she make her sides?

Explanation

Let's start by representing the length of the triangle, measured in inches, by the letter $x\text{.}$ That would also make the other side $x$ inches long.

Faven should now set up the Pythagorean theorem regarding the picture. That would be

\begin{equation*} x^2+x^2=16^2 \end{equation*}

Solving this equation, we have:

\begin{align*} x^2+x^2\amp=16^2\\ x^2+x^2\amp=256\\ 2x^2\amp=256\\ x^2\amp=128\\ \highlight{\sqrt{\unhighlight{x^2}}}\amp=\highlight{\sqrt{\unhighlight{128}}}\\ x\amp=\sqrt{\highlight{64}\cdot2}\\ x\amp=\sqrt{\highlight{64}}\cdot\sqrt{2}\\ x\amp=\highlight{8}\sqrt{2}\\ x\amp\approx11.3 \end{align*}

Faven should make the sides of her triangle about $11.3$ inches long to force the hypotenuse to be $16$ inches long.

In SectionÂ 8.4 we covered how to use the quadratic formula to solve any quadratic equation, as well as an algorithm to help solve linear and quadratic equations.

Solve the equations using the quadratic formula.

1. $x^2+4x=6$

2. $5x^2-2x+1=0$

Explanation
1. First we should change the equation into standard form.

\begin{align*} x^2+4x\amp=6\\ x^2+4x-6\amp=0 \end{align*}

Next, we check and see that we cannot factor the left side or use the square root property so we must use the quadratic formula. We identify that $\substitute{a=1}\text{,}$ $\substitute{b=4}\text{,}$ and $\substitute{c=-6}\text{.}$ We will substitute them into the quadratic formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{4}\pm\sqrt{(\substitute{4})^2-4(\substitute{1})(\substitute{-6})}}{2(\substitute{1})}\\ \amp=\frac{-4\pm\sqrt{16+24}}{2}\\ \amp=\frac{-4\pm\sqrt{40}}{2}\\ \amp=\frac{-4\pm\sqrt{\highlight{4}\cdot10}}{2}\\ \amp=\frac{-4\pm\sqrt{\highlight{4}}\cdot\sqrt{10}}{2}\\ \amp=\frac{-4\pm\highlight{2}\sqrt{10}}{2}\\ \amp=-\frac{4}{2}\pm\frac{2\sqrt{10}}{2}\\ \amp=-2\pm\sqrt{10} \end{align*}

So the solution set is $\left\{-2+\sqrt{10},-2-\sqrt{10}\right\}\text{.}$

2. Since the equation $5x^2-2x+1=0$ is already in standard form, we check and see that we cannot factor the left side or use the square root property so we must use the quadratic formula. We identify that $\substitute{a=5}\text{,}$ $\substitute{b=-2}\text{,}$ and $\substitute{c=1}\text{.}$ We will substitute them into the quadratic formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{(-2)}\pm\sqrt{\substitute{(-2)}^2-4(\substitute{5})(\substitute{1})}}{2(\substitute{5})}\\ \amp=\frac{2\pm\sqrt{4-20}}{10}\\ \amp=\frac{2\pm\sqrt{-16}}{10} \end{align*}

Since the solutions have square roots of negative numbers, we must conclude that there are no real solutions.

###### Example8.7.13Recognizing Linear and Quadratic Equations

Identify which equations are linear, which are quadratic, and which are neither.

1. $2(x-3)^2-5x=6$

2. $2(x-3)-5x=6$

3. $2x-6=7x^3$

4. $2x^2-6=7x^2$

5. $2\sqrt{x}-x-6=0$

6. $2x-(x-6)=0$

Explanation
1. $2(x-3)^2-5x=6$ is quadratic.

2. $2(x-3)-5x=6$ is linear.

3. $2x-6=7x^3$ is neither linear or quadratic.

4. $2x^2-6=7x^2$ is quadratic.

5. $2\sqrt{x}-x-6=0$ is neither linear or quadratic.

6. $2x-(x-6)=0$ is linear.

###### Example8.7.14Solving Linear and Quadratic Equations

Use 8.4.11 to help solve the equations after deciding if they are linear or quadratic.

1. $4x^2-11x+6=0$

2. $2(x-6)^2-16=0$

3. $2(x-6)-16=0$

4. $3(x-4)^2-15x=0$

Explanation
1. To solve the equation $4x^2-11x+6=0$ we first note that it is quadratic. Since there is a linear term ($-11x$), we must use either factoring or the quadratic formula, and we will try factoring first. Since the leading coefficient is $4\text{,}$ we will try the AC method. In this case, $ac=24\text{:}$ numbers that multiply to be $24$ and add to be $-11$ are $-8$ and $-3\text{.}$ So we split up th equation like this:

\begin{align*} 4x^2-11x+6\amp=0\\ 4x^2\mathbin{\highlight{-}}\mathbin{\highlight{8x-3x}}+6\amp=0\\ \left(4x^2\mathbin{\highlight{-}}\mathbin{\highlight{8x}}\right)+(\mathbin{\highlight{-}}\mathbin{\highlight{3x}}+6)\amp=0\\ 4x(x-2)-3(x-2)\amp=0\\ (x-2)(4x-3)\amp=0 \end{align*}
\begin{align*} x-2\amp=0 \amp \text{ or } \amp\amp 4x-3\amp=0\\ x\amp=2 \amp \text{ or } \amp\amp x\amp=\frac{3}{4} \end{align*}

So, the solution set is $\left\{2,\frac{3}{4}\right\}\text{.}$

2. To solve the equation $2(x-6)^2-16=0$ we first note that it is quadratic. Since there is no linear term, we should try using the square root method.

\begin{align*} 2(x-6)^2-16\amp=0\\ 2(x-6)^2\amp=16\\ (x-6)^2\amp=8 \end{align*}
\begin{align*} x-6\amp=\sqrt{8} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{8}\\ x-6\amp=\sqrt{\highlight{4}\cdot2} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{\highlight{4}\cdot2}\\ x-6\amp=\sqrt{\highlight{4}}\cdot\sqrt{2} \amp \text{ or } \amp\amp x-6\amp=-\sqrt{\highlight{4}}\cdot\sqrt{2}\\ x-6\amp=\highlight{2}\sqrt{2} \amp \text{ or } \amp\amp x-6\amp=-\highlight{2}\sqrt{2}\\ x\amp=6+\highlight{2}\sqrt{2} \amp \text{ or } \amp\amp x\amp=6-\highlight{2}\sqrt{2} \end{align*}

So, the solution set is $\left\{6+2\sqrt{2},6-2\sqrt{2}\right\}\text{.}$

3. To solve the equation $2(x-6)-16=0$ we first we first note that it is linear. Since it is linear, we just need to isolate the terms with the variable on one side and all the other terms on the other side of the equals sign.

\begin{gather*} 2(x-6)-16=0\\ 2x-12-16=0\\ 2x-28=0\\ 2x=28\\ x=14 \end{gather*}

So, the solution set is $\{14\}\text{.}$

4. To solve the equation $2(x-6)^2-15x=0$ we first note that it is quadratic. Since there is a linear term, we must use either factoring or the quadratic formula. Before we can decide which to use, we need to put the equation in standard form:

\begin{align*} 3(x-4)^2-15x=0\\ 3(x-4)(x-4)-15x=0\\ 3(x^2-8x+16)-15x=0\\ 3x^2-24x+48-15x=0\\ 3x^2-39x+48=0\\ \end{align*}

Now we can see that the left hand side does not factor easily, so we will fall back on the quadratic formula. We identify that $\substitute{a=3}\text{,}$ $\substitute{b=-39}\text{,}$ and $\substitute{c=48}\text{.}$

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-\substitute{(-39)}\pm\sqrt{\substitute{(-39)}^2-4(\substitute{3})(\substitute{48})}}{2(\substitute{3})}\\ \amp=\frac{39\pm\sqrt{1521-576}}{6}\\ \amp=\frac{39\pm\sqrt{945}}{6}\\ \amp=\frac{39\pm\sqrt{\highlight{9}\cdot105}}{6}\\ \amp=\frac{39\pm\sqrt{\highlight{9}}\cdot\sqrt{105}}{6}\\ \amp=\frac{39\pm3\sqrt{105}}{6} \end{align*}

So the solution set is $\left\{\frac{39+3\sqrt{105}}{6},\frac{39-3\sqrt{105}}{6}\right\}\text{.}$

### Subsection8.7.5Complex Solutions to Quadratic Equations

In SectionÂ 8.5 we covered what both imaginary numbers and complex numbers are, as well as how to solve quadratic equations where the solutions are imaginary numbers or complex numbers.

###### Example8.7.15Imaginary Numbers

Simplify the expression $\sqrt{-12}$ using the imaginary number, $i\text{.}$

Explanation

Start by splitting the $-1$ from the $12$ and by looking for the largest perfect-square factor of $-12\text{,}$ which happens to be $4\text{.}$

\begin{align*} \sqrt{-12}\amp=\sqrt{\highlight{4}\cdot\lighthigh{-1}\cdot3}\\ \amp=\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{3}\\ \amp=\highlight{2}\lighthigh{i}\sqrt{3} \end{align*}
###### Example8.7.16Solving Quadratic Equations with Imaginary Solutions

Solve for $m$ in $2m^2+16=0\text{,}$ where $p$ is an imaginary number.

Explanation

There is no $m$ term so we will use the square root method.

\begin{align*} 2m^2+16\amp=0\\ 2m^2\amp=-16\\ m^2\amp=-8 \end{align*}
\begin{align*} m\amp=-\sqrt{-8} \amp\text{ or }\amp\amp m\amp=\sqrt{-8}\\ m\amp=-\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2} \amp\text{ or }\amp\amp m\amp=\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2}\\ m\amp=-\highlight{2}\lighthigh{i}\sqrt{2}\amp\text{ or }\amp\amp m\amp=\highlight{2}\lighthigh{i}\sqrt{2} \end{align*}

The solution set is $\left\{-\highlight{2}\lighthigh{i}\sqrt{2},\highlight{2}\lighthigh{i}\sqrt{2}\right\}\text{.}$

###### Example8.7.17Solving Quadratic Equations with Complex Solutions

Solve the equation $3(v-2)^2+54=0\text{,}$ where $v$ is a complex number.

Explanation
\begin{align*} 3(v-2)^2+54\amp=0\\ 3(v-2)^2\amp=-54\\ (v-2)^2\amp=-18 \end{align*}
\begin{align*} v-2\amp=-\sqrt{-18}\amp\text{ or }\amp\amp v-2\amp=\sqrt{-18}\\ v-2\amp=-\sqrt{\highlight{9}\cdot\lighthigh{-1}\cdot2}\amp\text{ or }\amp\amp v-2\amp=\sqrt{\highlight{9}\cdot\lighthigh{-1}\cdot2}\\ v-2\amp=-\sqrt{\highlight{9}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2}\amp\text{ or }\amp\amp v-2\amp=\sqrt{\highlight{9}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{2}\\ v-2\amp=-\highlight{3}\lighthigh{i}\sqrt{2}\amp\text{ or }\amp\amp v-2\amp=\highlight{3}\lighthigh{i}\sqrt{2}\\ v\amp=2-\highlight{3}\lighthigh{i}\sqrt{2}\amp\text{ or }\amp\amp v\amp=2+\highlight{3}\lighthigh{i}\sqrt{2} \end{align*}

So, the solution set is $\left\{2+\highlight{3}\lighthigh{i}\sqrt{2},2-\highlight{3}\lighthigh{i}\sqrt{2}\right\}\text{.}$

### Subsection8.7.6Strategies for Solving Quadratic Equations

In SectionÂ 8.6 we reviews all three methods for solving quadratic equations that we know. For the full explanation of solving using the factoring, visit SectionÂ 8.1, solving using the square root method, visit SectionÂ 8.3, and for more on the quadratic formula, visit SectionÂ 8.4.

###### Example8.7.18How to Choose a Method for Solving a Quadratic Equation

Solve the quadratic equations using an effective method.

1. $(x-4)^2-2=0$

2. $(x-4)^2-2x=0$

3. $(x-4)^2+2x=0$

Explanation

All three of the equations here are very similar, so we will need to examine them closely to choose the best method for solving them.

1. To solve the equation $(x-4)^2-2=0\text{,}$ first note that there is no linear term: there is only a square and a constant. This leads us to consider the square root method. Before doing that, isolate the square:

\begin{align*} (x-4)^2-2\amp=0\\ (x-4)^2\amp=2 \end{align*}

Now we can apply the square root method to the equation.

\begin{align*} x-4\amp=\sqrt{2}\amp\text{ or }\amp\amp x-4\amp=-\sqrt{2}\\ x\amp=4+\sqrt{2}\amp\text{ or }\amp\amp x\amp=4-\sqrt{2} \end{align*}

So the solution set is $\left\{4+\sqrt{2},4-\sqrt{2}\right\}$

2. To solve the equation $(x-4)^2-2x=0\text{,}$ first note that there is a linear term ($-2x$), so we must use either factoring or the quadratic formula. To use either, we must first put the equation in standard form.

\begin{align*} (x-4)^2-2x\amp=0\\ (x-4)(x-4)-2x\amp=0\\ x^2-8x+16-2x\amp=0\\ x^2-10x+16\amp=0 \end{align*}

Now that the equation is in standard form, we can decide whether to use factoring or the quadratic formula. While the quadratic formula always works, it can take more time than factoring if factoring is possible. In this case, factoring entails answering the question âare there two integers that multiply to be $16$ and add to be $-10\text{?}$â The answer is âyesâ: $-8$ and $-2$ are such numbers.

\begin{align*} x^2-10x+16\amp=0\\ (x-8)(x-2)\amp=0 \end{align*}
\begin{align*} x-8\amp=0\amp\text{ or }\amp\amp x-2\amp=0\\ x\amp=8\amp\text{ or }\amp\amp x\amp=2 \end{align*}

So the solution set is $\{2,8\}\text{.}$

3. To solve the equation $(x-4)^2+2x=0\text{,}$ first note that there is a linear term ($+2x$), so we must use either factoring or the quadratic formula. To use either, we must first put the equation in standard form.

\begin{align*} (x-4)^2+2x\amp=0\\ (x-4)(x-4)+2x\amp=0\\ x^2-8x+16+2x\amp=0\\ x^2-6x+16\amp=0 \end{align*}

Now that the equation is in standard form, we can decide whether to use factoring or the quadratic formula. In this case, factoring entails answering the question âare there two integers that multiply to be $16$ and add to be $-6\text{?}$â The answer is âno,â so we must use the quadratic formula. First, identify that $\substitute{a=1}\text{,}$ $\substitute{b=-6}\text{,}$ and $\substitute{c=16}\text{.}$

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{-6})\pm\sqrt{(\substitute{-6})^2-4(\substitute{1})(\substitute{16})}}{2(\substitute{1})}\\ \amp=\frac{6\pm\sqrt{36-48}}{2}\\ \amp=\frac{6\pm\sqrt{-12}}{2}\\ \end{align*}

At this point, we notice that the solutions are complex. Continue to simplify until they are completely reduced.

\begin{align*} x\amp=\frac{6\pm\sqrt{\highlight{4}\cdot\lighthigh{-1}\cdot3}}{2}\\ \amp=\frac{6\pm\sqrt{\highlight{4}}\cdot\sqrt{\lighthigh{-1}}\cdot\sqrt{3}}{2}\\ \amp=\frac{6\pm\highlight{2}\lighthigh{i}\sqrt{3}}{2}\\ \amp=\frac{6}{2}\pm\frac{2i\sqrt{3}}{2}\\ \amp=3\pm i\sqrt{3} \end{align*}

So the solution set is $\left\{3+i\sqrt{3},3-i\sqrt{3}\right\}\text{.}$

### Subsection8.7.7Exercises

###### 1

Solve the equation.

${x^{2}-6x-27} = 0$

###### 2

Solve the equation.

${x^{2}-5x-50} = 0$

###### 3

Solve the equation.

${x^{2}+11x} = {-18}$

###### 4

Solve the equation.

${x^{2}+18x} = {-80}$

###### 5

Solve the equation.

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

###### 6

Solve the equation.

${x^{2}} = {7x}$

###### 7

Solve the equation.

${x^{2}-8x+16}=0$

###### 8

Solve the equation.

${x^{2}-10x+25}=0$

###### 9

Solve the equation.

${4x^{2}}={-41x-10}$

###### 10

Solve the equation.

${4x^{2}}={-25x-36}$

###### 11

Solve the equation.

${x\!\left(x+20\right)}={5\!\left(2x-5\right)}$

###### 12

Solve the equation.

${x\!\left(x+36\right)}={9\!\left(2x-9\right)}$

###### 13

A rectangleâs base is ${7\ {\rm in}}$ shorter than four times its height. The rectangleâs area is ${65\ {\rm in^{2}}}\text{.}$ Find this rectangleâs dimensions.

The rectangleâs height is .

The rectangleâs base is .

###### 14

A rectangleâs base is ${7\ {\rm in}}$ shorter than four times its height. The rectangleâs area is ${2\ {\rm in^{2}}}\text{.}$ Find this rectangleâs dimensions.

The rectangleâs height is .

The rectangleâs base is .

###### 15

Without using a calculator, estimate the value of $\sqrt{22}\text{:}$

• 5.69

• 4.69

• 4.31

• 5.31

###### 16

Without using a calculator, estimate the value of $\sqrt{38}\text{:}$

• 6.16

• 6.84

• 5.84

• 5.16

###### 17

Evaluate the following.

$\displaystyle{\sqrt{{{\frac{25}{81}}}}={}}$.

###### 18

Evaluate the following.

$\displaystyle{\sqrt{{{\frac{36}{49}}}}={}}$.

###### 19

Evaluate the following.

$\sqrt{-64}=$.

###### 20

Evaluate the following.

$\sqrt{-100}=$.

###### 21

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ \frac{{\sqrt{6}}}{{\sqrt{216}}} =}$

###### 22

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ \frac{{\sqrt{4}}}{{\sqrt{144}}} =}$

###### 23

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ {\sqrt{8}} = }$

###### 24

Simplify the radical expression or state that it is not a real number.

$\displaystyle{ {\sqrt{147}} = }$

###### 25

Simplify the expression.

$\displaystyle{3\sqrt{5} \cdot 3\sqrt{40}=}$

###### 26

Simplify the expression.

$\displaystyle{3\sqrt{15} \cdot 2\sqrt{30}=}$

###### 27

Simplify the expression.

$\displaystyle{ {\sqrt{\frac{28}{13}}} \cdot {\sqrt{\frac{4}{13}}} =}$

###### 28

Simplify the expression.

$\displaystyle{ {\sqrt{\frac{15}{19}}} \cdot {\sqrt{\frac{5}{19}}} =}$

###### 29

Simplify the expression.

$\displaystyle{{\sqrt{20}} + {\sqrt{45}} =}$

###### 30

Simplify the expression.

$\displaystyle{{\sqrt{175}} + {\sqrt{112}} =}$

###### 31

Simplify the expression.

$\displaystyle{{\sqrt{176}} - {\sqrt{44}} =}$

###### 32

Simplify the expression.

$\displaystyle{{\sqrt{8}} - {\sqrt{32}} =}$

###### 33

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{\sqrt{3}} = }$

###### 34

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{\sqrt{3}} = }$

###### 35

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{2}{5\sqrt{5}} = }$

###### 36

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{7}{3\sqrt{6}} = }$

###### 37

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{{\sqrt{75}}} = }$

###### 38

Rationalize the denominator and simplify the expression.

$\displaystyle{ \frac{1}{{\sqrt{175}}} = }$

###### 39

Expand and simplify the expression.

$\displaystyle{\left(7 - {\sqrt{7}}\right)\left(10 - 3 {\sqrt{7}}\right) =}$

###### 40

Expand and simplify the expression.

$\displaystyle{\left(5 - {\sqrt{7}}\right)\left(8 - 5 {\sqrt{7}}\right) =}$

###### 41

Expand and simplify the expression.

$\displaystyle{\left({\sqrt{5}} + {\sqrt{13}}\right)\left({\sqrt{5}} - {\sqrt{13}}\right) =}$

###### 42

Expand and simplify the expression.

$\displaystyle{\left({\sqrt{5}} + {\sqrt{7}}\right)\left({\sqrt{5}} - {\sqrt{7}}\right) =}$

###### 43

Solve the equation.

$x^2 = 20$

###### 44

Solve the equation.

$x^2 = 99$

###### 45

Solve the equation.

$144x^2 = 49$

###### 46

Solve the equation.

$16x^2 = 9$

###### 47

Solve the equation.

$\left(x+5\right)^2 = 121$

###### 48

Solve the equation.

$\left(x+7\right)^2 = 49$

###### 49

Solve the equation.

$2 - 2 ( t - 5 )^2 = -6$

###### 50

Solve the equation.

$54 - 5 ( x - 5 )^2 = 9$

###### 51

Find the value of $x\text{.}$

$x={}$

###### 52

Find the value of $x\text{.}$

$x={}$

###### 53

Parnell is designing a rectangular garden. The gardenâs diagonal must be $2$ feet, and the ratio between the gardenâs base and height must be $4:3\text{.}$ Find the length of the gardenâs base and height.

The gardenâs base is feet and its height is .

###### 54

Gregory is designing a rectangular garden. The gardenâs diagonal must be $42.9$ feet, and the ratio between the gardenâs base and height must be $12:5\text{.}$ Find the length of the gardenâs base and height.

The gardenâs base is feet and its height is .

###### 55

Solve the equation.

${14x^{2}-11x-9}=0$

###### 56

Solve the equation.

${18x^{2}-37x-20}=0$

###### 57

Solve the equation.

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

###### 58

Solve the equation.

${x^{2}}= {5x-3}$

###### 59

Solve the equation.

${2x^{2}+4x+5}= 0$

###### 60

Solve the equation.

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

###### 61

Solve the equation.

${10t+7} = {t+88}$

###### 62

Solve the equation.

${9b+10} = {b+66}$

###### 63

Solve the equation.

$-1-3y^2 = -3$

###### 64

Solve the equation.

$-8-7r^2 = -10$

###### 65

Solve the equation.

$x^2+39x= 0$

###### 66

Solve the equation.

$x^2+59x= 0$

###### 67

Solve the equation.

${x^{2}+8x} = {9}$

###### 68

Solve the equation.

${x^{2}-3x} = {54}$

###### 69

Solve the equation.

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

###### 70

Solve the equation.

$\left(x - 5\right)^2 = 121$

###### 71

Solve the equation.

${x^{2}}= {-9x-19}$

###### 72

Solve the equation.

${x^{2}}= {-7x-8}$

###### 73

An object is launched upward at the height of $320$ meters. Itâs height can be modeled by

\begin{equation*} h=-4.9t^2+90t+320\text{,} \end{equation*}

where $h$ stands for the objectâs height in meters, and $t$ stands for time passed in seconds since its launch. The objectâs height will be $350$ meters twice before it hits the ground. Find how many seconds since the launch would the objectâs height be $350$ meters. Round your answers to two decimal places if needed.

The objectâs height would be $350$ meters the first time at seconds, and then the second time at seconds.

###### 74

An object is launched upward at the height of $340$ meters. Itâs height can be modeled by

\begin{equation*} h=-4.9t^2+70t+340\text{,} \end{equation*}

where $h$ stands for the objectâs height in meters, and $t$ stands for time passed in seconds since its launch. The objectâs height will be $350$ meters twice before it hits the ground. Find how many seconds since the launch would the objectâs height be $350$ meters. Round your answers to two decimal places if needed.

The objectâs height would be $350$ meters the first time at seconds, and then the second time at seconds.

###### 75

Simplify the radical and write it into a complex number.

$\displaystyle{ \sqrt{-48} =}$

###### 76

Simplify the radical and write it into a complex number.

$\displaystyle{ \sqrt{-98} =}$

###### 77

Solve the quadratic equation. Solutions could be complex numbers.

${-2x^{2}} - 4 = 8$

###### 78

Solve the quadratic equation. Solutions could be complex numbers.

${4x^{2}} + 7 = -5$

###### 79

Solve the quadratic equation. Solutions could be complex numbers.

$-9(y - 4)^2+8 = 584$

###### 80

Solve the quadratic equation. Solutions could be complex numbers.

$7(y+7)^2+8 = -167$

###### 81

Solve the equation.

$2x^2 + 29= 0$

###### 82

Solve the equation.

$41x^2 + 37= 0$

###### 83

Solve the equation.

${5x^{2}}={-31x-44}$

###### 84

Solve the equation.

${5x^{2}}={-27x-36}$

###### 85

Solve the equation.

${x^{2}+4x+1}= 0$

###### 86

Solve the equation.

${x^{2}+10x+7}= 0$

###### 87

Solve the equation.

$28 - 6 ( x+6 )^2 = 4$

###### 88

Solve the equation.

$34 - 4 ( y+6 )^2 = -2$

###### 89

Solve the equation.

${x^{2}+7x} = {-12}$

###### 90

Solve the equation.

${x^{2}+14x} = {-45}$