Back to Solve rational and radical equations — Problem 2 · Task Set 33

HSA.REI.A.2 Tutor Intake -- Rational and Radical Equations

This short check helps your tutor understand where to start. Answer each
question without help. If you are not sure, give your best try — the goal
is to find what to work on together, not to grade you.

Grade 9·8 problems·~13 min·Common Core Math - HS Algebra·standard·hsa-rei-a-2

Work through problems with immediate feedback

Concepts

A student solves a radical equation and obtains $x = 4$ and $x = -1$ .
After checking in the original equation:

$x = 4$ : left side $= 3$ , right side $= 3$ ✓
$x = -1$ : left side $= 2$ , right side $= -1$ ✗

What is the correct solution set?

$\{4, -1\}$ — both values are solutions because the algebra produced them.

$\{4\}$ — only $x = 4$ satisfies the original equation; $x = -1$ is extraneous.

$\{-1\}$ — only the negative solution is valid for radical equations.

$\emptyset$ — when an extraneous solution appears, the equation has no solution.