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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 8: Q. 8.123 (page 817)

Solve: $2 \sqrt{4 a + 4} - 16 = 16$

Short Answer

Expert verified

The solution is $a = 63$ .

Step by step solution

01

Step 1. Isolate the radical

In the equation, isolate the radical by first adding $16$ both sides.

localid="1644860736246" $\begin{array}{rcl} 2 \sqrt{4 a + 4} - 16 & = & 16 \\ 2 \sqrt{4 a + 4} - 16 + 16 & = & 16 + 16 \\ 2 \sqrt{4 a + 4} & = & 32 \end{array}$

Now divide both sides by $2$

$\begin{array}{rcl} \frac{2 \sqrt{4 a + 4}}{2} & = & \frac{32}{2} \\ \sqrt{4 a + 4} & = & 16 \end{array}$

02

Step 2. Square both sides and solve for a

Remove the radical by squaring both sides.

$\begin{array}{rcl} {(\sqrt{4 a + 4})}^{2} & = & 16^{2} \\ 4 a + 4 & = & 256 \end{array}$

Subtract $4$ from both sides

$\begin{array}{rcl} 4 a + 4 - 4 & = & 256 - 4 \\ 4 a & = & 252 \end{array}$

Now divide both sides by $4$ and solve for $a$ .

$\begin{array}{rcl} \frac{4 a}{4} & = & \frac{252}{4} \\ a & = & 63 \end{array}$

03

Step 3. Check the solution

Substitute $63$ for $a$ in the given equation.

$\begin{array}{rcl} 2 \sqrt{4 a + 4} - 16 & = & 16 \\ 2 \sqrt{4 路 63 + 4} - 16 & = & 16 \\ 2 \sqrt{252 + 4} - 16 & = & 16 \\ 2 \sqrt{256} - 16 & = & 16 \\ 2 路 16 - 16 & = & 16 \\ 32 - 16 & = & 16 \\ 16 & = & 16 \end{array}$

So, we get a true statement. Thus the solution found is correct.

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Most popular questions from this chapter

In the following exercises, write with a rational exponent.

$鈸� \sqrt[7]{x} 鈸� \sqrt[9]{y} 鈸� \sqrt[5]{f}$

Simplify:

$(a) \sqrt{180 m^{9} n^{11}} (b) \sqrt{72 x^{6} y^{5}} (c) \sqrt{80 x^{7} y^{4}} .$

In the following exercises, simplify using absolute value signs as needed. $\sqrt{147 m^{7} n^{11}}, \sqrt[3]{48 x^{6} y^{7}}, \sqrt[4]{32 x^{5} y^{4}}$

In the following exercises, use the Product Property to simplify radical expressions $\sqrt{147}$

In the following exercises, solve.

$\sqrt[3]{2 x} = - 2$

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