Checkpoint: Solving Radical Equations Solution

Solve. Check your work.

$\sqrt{x + 1} = 5 - \sqrt{x}$

$\begin{array}{rcl} {(\sqrt{x + 1})}^{2} & = & {(5 - \sqrt{x})}^{2} \\ x + 1 & = & (5 - \sqrt{x}) (5 - \sqrt{x}) \\ x + 1 & = & 25 - 10 \sqrt{x} + x \\ - 24 & = & - 10 \sqrt{x} \\ {(\frac{12}{5})}^{2} & = & {(\sqrt{x})}^{2} \\ x & = & \frac{144}{25} \end{array}$

Check

$\begin{array}{rcl} \frac{144}{25} & = & 5.76 \\ \sqrt{5.76 + 1} & = & 5 - \sqrt{5.76} \\ \sqrt{6.76} & = & 5 - 2.4 \\ 2.6 & = & 2.6 ✓ \end{array}$

Note

Q: How do you know that the value is not extraneous?

A: Use substitution and determine if both sides are equivalent.

Q: What would happen if the right side of the equation was addition rather than subtraction?

A: The equation would have no solution because the sides would not be equal. (Hint: Use the check to see how this would change the problem.)

Explaining each step out loud or to an instructor can be helpful in remembering to do the same thing on both sides of the equation.