Solving radical equations; extraneous solutions

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $\sqrt{x} = -3$ . Squaring gives $x = 9$ . Does 9 work?

No, $\sqrt{9} = 3$
Yes

Key idea

Undo the root, then check.

To solve a radical equation, isolate the radical, raise both sides to the matching power (square for

\sqrt{\ }

, cube for

\sqrt[3]{\ }

), and solve. Squaring can create extraneous solutions: always check in the original equation. Cubing does not.

Worked example

$\sqrt{x + 6} = x$ .

x + 6 = x^2

Square.

x^2 - x - 6 = 0

, so

x = 3

x = -2

Solve.

Only

x = 3

√9 = 3 ✓; √4 = 2 ≠ −2 ✗.

Your turn, guided

Your turn: $\sqrt{x - 1} = 4$ .

x = 17

x − 1 = 16.

Practice ★

Solve $\sqrt{x + 2} = 8$ .
Solve $\sqrt{x + 8} = 7$ .

Solve $\sqrt[3]{x - 8} = 3$ .

Watch out!

Practice ★★

Solve $\sqrt{x + 6} = x$ . (Squaring gives two candidates; enter the one that works.)
Solve $\sqrt{x + 30} = x$ . (Squaring gives two candidates; enter the one that works.)

Exit ticket

Three questions to check you have got it.

Solve $\sqrt{x + 2} = 8$ .
Solve $\sqrt{x + 8} = 7$ .
Solve $\sqrt{x + 4} = 6$ .

Summary card

Radical equations

Isolate → square (or cube) → solve → check.

Watch out: squaring can add extraneous solutions.

End of the lesson

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