Solving Radical Equations

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Algebra 2

Radical Functions

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Solving Radical Equations 1.7 - Solution

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a

First, we need to undo the radical expressions by squaring both sides of the equation.

$\sqrt{x+4}=\sqrt{7-x}$

RaiseEqn$\text{LHS}^{2}=\text{RHS}^{2}$

$x+4=7-x$

AddEqn$\text{LHS}+x=\text{RHS}+x$

$2x+4=7$

SubEqn$\text{LHS}-4=\text{RHS}-4$

$2x=3$

DivEqn$\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.$

$x=1.5$

Next we need to test $x=1.5$ to ensure that it is not an extraneous solution.

$\sqrt{x+4}=\sqrt{7-x}$

Substitute$x={\color{#0000FF}{1.5}}$

$\sqrt{{\color{#0000FF}{1.5}}+4}\stackrel{?}{=}\sqrt{7-{\color{#0000FF}{1.5}}}$

AddSubTermsAdd and subtract terms

$\sqrt{5.5}=\sqrt{5.5}$

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.

$\sqrt{5x-16}=\sqrt{2x+2}$

RaiseEqn$\text{LHS}^{2}=\text{RHS}^{2}$

$5x-16=2x+2$

SubEqn$\text{LHS}-2x=\text{RHS}-2x$

$3x-16=2$

AddEqn$\text{LHS}+16=\text{RHS}+16$

$3x=18$

DivEqn$\left.\text{LHS}\middle/3\right.=\left.\text{RHS}\middle/3\right.$

$x=6$

We arrived at $x=6.$ Let's test it by substituting $6$ for $x$ in the original equation.

$\sqrt{5x-16}=\sqrt{2x+2}$

Substitute$x={\color{#0000FF}{6}}$

$\sqrt{5 \cdot {\color{#0000FF}{6}}-16}\stackrel{?}{=}\sqrt{2 \cdot {\color{#0000FF}{6}}+2}$

MultiplyMultiply

$\sqrt{30-16}\stackrel{?}{=}\sqrt{12+2}$

AddSubTermsAdd and subtract terms

$\sqrt{14}=\sqrt{14}$

This result shows that $x=6$ is a solution to the equation.