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Solving Radical Equations

Solving Radical Equations 1.7 - Solution

arrow_back Return to Solving Radical Equations
a
First, we need to undo the radical expressions by squaring both sides of the equation.
x+4=7x\sqrt{x+4}=\sqrt{7-x}
x+4=7xx+4=7-x
2x+4=72x+4=7
2x=32x=3
x=1.5x=1.5
Next we need to test x=1.5x=1.5 to ensure that it is not an extraneous solution.
x+4=7x\sqrt{x+4}=\sqrt{7-x}
1.5+4=?71.5\sqrt{{\color{#0000FF}{1.5}}+4}\stackrel{?}{=}\sqrt{7-{\color{#0000FF}{1.5}}}
5.5=5.5\sqrt{5.5}=\sqrt{5.5}
Since we arrived at a true statement we know that x=1.5x=1.5 is a solution to the equation.
b
We use the same method here as in Part A.
5x16=2x+2\sqrt{5x-16}=\sqrt{2x+2}
5x16=2x+25x-16=2x+2
3x16=23x-16=2
3x=183x=18
x=6x=6
We arrived at x=6.x=6. Let's test it by substituting 66 for xx in the original equation.
5x16=2x+2\sqrt{5x-16}=\sqrt{2x+2}
5616=?26+2\sqrt{5 \cdot {\color{#0000FF}{6}}-16}\stackrel{?}{=}\sqrt{2 \cdot {\color{#0000FF}{6}}+2}
3016=?12+2\sqrt{30-16}\stackrel{?}{=}\sqrt{12+2}
14=14\sqrt{14}=\sqrt{14}
This result shows that x=6x=6 is a solution to the equation.