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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. 8.128 (page 819)

Solve: $\sqrt{x} + 2 = \sqrt{x + 16}$

Short Answer

Expert verified

The solution is $x = 9$ .

Step by step solution

01

Step 1. Square both sides

Consider the equation $\sqrt{x} + 2 = \sqrt{x + 16}$ .

The radical on the right is isolated. Now we square both sides and simplify.

role="math" localid="1644863696250"

\begin{array}{rcl} {(\sqrt{x} + 2)}^{2} & = & {(\sqrt{x + 16})}^{2} \\ {(\sqrt{x})}^{2} + 2 路 \sqrt{x} 路 2 + 2^{2} & = & x + 16 \\ x + 4 \sqrt{x} + 4 & = & x + 16 \end{array}

02

Step 2. Isolate the radical and solve for x

Subtract $x + 4$ from both sides

$\begin{array}{rcl} x + 4 \sqrt{x} + 4 - (x + 4) & = & x + 16 - (x + 4) \\ x + 4 \sqrt{x} + 4 - x - 4 & = & x + 16 - x - 4 \\ 4 \sqrt{x} & = & 12 \end{array}$

Divide both side by $4$ .

$\begin{array}{rcl} \frac{4 \sqrt{x}}{4} & = & \frac{12}{4} \\ \sqrt{x} & = & 3 \end{array}$

Now square both sides and solve the equation.

$\begin{array}{rcl} {(\sqrt{x})}^{2} & = & 3^{2} \\ x & = & 9 \end{array}$

03

Step 3. Check the solution

Substitute $9$ for $x$ in the original equation.

$\begin{array}{rcl} \sqrt{x} + 2 & = & \sqrt{x + 16} \\ \sqrt{9} + 2 & = & \sqrt{9 + 16} \\ 3 + 2 & = & \sqrt{25} \\ 5 & = & 5 \end{array}$

As the statement is true, so the found solution is correct.

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Most popular questions from this chapter

In the following exercises, simplify using absolute value signs as needed. $\sqrt{96 r^{3} s^{3}}, \sqrt[3]{80 x^{7} y^{6}}, \sqrt[4]{80 x^{8} y^{9}}$

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