Week 5 – Solving Radical Equations

Consider the equation $\sqrt{3x+1}=5$

To solve the equation, we need to isolate x. Square roots and squaring are inverse operations. So, square both sides of the equation to “undo” the square root.

$(\sqrt{3x+1})^2=(5)^2$ ,

$3x+1=25$

Now, we subtract 1 from each side to get rid of the 1 added to the equation. Remember; what you do to one side, you have to do to the other.

$3x=24$

To isolate x, divide both sides by 3.

$\frac{3x}{3}= \frac{24}{3}$ ,

$x=8$

Check the answer. Substitute x = 8 in the original equation.

$\sqrt{3(8)+1}=5$ ,

$\sqrt{24+1}=5$ ,

$\sqrt{25}=5$ ,

$5=5$

Since the left side equals the right side, the root of the equation is 5.

Restriction: $3x+1>0$

$3x>-1$

$\frac{3x}{3}> \frac{-1}{3}$ ,

$x > \frac{-1}{3}$ or $x = \frac{-1}{3}$

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Week 5 – Solving Radical Equations

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