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## Solving Radical Equations 1.7 - Solution

a
First, we need to undo the radical expressions by squaring both sides of the equation.
$\sqrt{x+4}=\sqrt{7-x}$
$x+4=7-x$
$2x+4=7$
$2x=3$
$x=1.5$
Next we need to test $x=1.5$ to ensure that it is not an extraneous solution.
$\sqrt{x+4}=\sqrt{7-x}$
$\sqrt{{\color{#0000FF}{1.5}}+4}\stackrel{?}{=}\sqrt{7-{\color{#0000FF}{1.5}}}$
$\sqrt{5.5}=\sqrt{5.5}$
Since we arrived at a true statement we know that $x=1.5$ is a solution to the equation.
b
We use the same method here as in Part A.
$\sqrt{5x-16}=\sqrt{2x+2}$
$5x-16=2x+2$
$3x-16=2$
$3x=18$
$x=6$
We arrived at $x=6.$ Let's test it by substituting $6$ for $x$ in the original equation.
$\sqrt{5x-16}=\sqrt{2x+2}$
$\sqrt{5 \cdot {\color{#0000FF}{6}}-16}\stackrel{?}{=}\sqrt{2 \cdot {\color{#0000FF}{6}}+2}$
$\sqrt{30-16}\stackrel{?}{=}\sqrt{12+2}$
$\sqrt{14}=\sqrt{14}$
This result shows that $x=6$ is a solution to the equation.