Solve Radical Equations

Solving radical equations. In MathBound, the free Algebra I adventure game, Stump Puller asks these problems, and the only way past is to solve them.

Part of Radicals and More Functions: chapter 11 of the standard course.

10 questions, no sign-up, nothing saved. Climbs from the basics to harder questions.

What you will practice

Isolate and Solve. Solve a basic equation where the square root is already isolated by squaring both sides.
Check Extraneous Solutions. Identify and reject solutions that arise from squaring but make the original equation false (e.g. root equals negative).
Solve with Quadratics. Solve an equation where squaring both sides yields a quadratic equation that must be factored.
Solve with Two Radicals. Solve an equation with a square root on both sides by squaring both sides to eliminate the radicals simultaneously.

Made by the game’s own question engine. Try each one, then open its steps.

Solve:

x = \frac{1}{4}

Show the worked steps

Answer: $x = \frac{1}{16}$

Solve:

4 y - 4 = - 4

Show the worked steps

$The$ $square$ $root$ $on$ $the$ $left$ $can$ $never$ $be$ $negative.$
$It$ $is$ $set$ $equal$ $to$ $- 4,$ $so$ $there$ $is$ $no$ $solution.$
$Squaring$ $would$ $hide$ $this$ $and$ $produce$ $a$ $value$ $that$ $fails$ $the$ $original$ $equation.$

Answer: $No$ $solution$

Solve:

x + 12 = x

Show the worked steps

$Square$ $both$ $sides:$ $x + 12 = x^{2} .$
$x^{2} - x - 12 = 0$ $factors,$ $giving$ $x = 4$ $and$ $x = - 3 .$
$The$ $negative$ $root$ $fails$ $the$ $original$ $equation,$ $so$ $only$ $x = 4$ $works.$

Answer: $x = 4$

Textbook	Lessons
Standard course	11.3
Big Ideas Math Algebra 1	10.3
Big Ideas Math Algebra 1 (2022)	10.3
CPM Core Connections Algebra	10.2.3
Eureka Math Algebra I (EngageNY)	1.12-1.13
Glencoe Algebra 1	10-4
OpenStax Elementary Algebra 2e	9.6
Pearson Algebra 1 Common Core	10-4