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Algebra II

Solving Radical Equations

Lesson

A radical equation has a variable under a square root (or other root). The big move: isolate the radical, then square both sides.

Three-step recipe

  1. Get the radical alone on one side.
  2. Square both sides to eliminate the root.
  3. Solve the resulting equation — and always check.

Worked example 1

x+4=5\sqrt{x + 4} = 5

Square both sides:

x+4=25  ⟹  x=21x + 4 = 25 \implies x = 21

Check: 21+4=25=5\sqrt{21 + 4} = \sqrt{25} = 5 ✓

Worked example 2 — isolate first

x+3=8\sqrt{x} + 3 = 8

Subtract 3, then square:

x=5  ⟹  x=25\sqrt{x} = 5 \implies x = 25

Watch for extraneous solutions

Squaring can introduce values that don’t actually solve the original equation. ALWAYS check by plugging back in.

Example: x=x−2\sqrt{x} = x - 2. Squaring gives x=x2−4x+4x = x^2 - 4x + 4, so x=1x = 1 or x=4x = 4. But 1=1\sqrt{1} = 1 while 1−2=−11 - 2 = -1 — that’s extraneous. Only x=4x = 4 works.

No-solution flag

If the radical alone equals a NEGATIVE number, there is no solution. A square root is never negative.

Example: x=−3\sqrt{x} = -3 — no real solution.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

x=4\sqrt{x} = 4

Problem 2

x=5\sqrt{x} = 5

Problem 3

x−1=3\sqrt{x - 1} = 3

Problem 4

x+4=2\sqrt{x + 4} = 2

Practice

Standard problems matching the lesson.

Problem 5

x=6\sqrt{x} = 6

Problem 6

x=0\sqrt{x} = 0

Problem 7

x−3=5\sqrt{x - 3} = 5

Problem 8

x+7=4\sqrt{x + 7} = 4

Problem 9

2x=4\sqrt{2x} = 4

Problem 10

3x=6\sqrt{3x} = 6

Problem 11

x+5=7\sqrt{x + 5} = 7

Problem 12

2x+1=3\sqrt{2x + 1} = 3

Problem 13

4x=10\sqrt{4x} = 10

Problem 14

x+3=8\sqrt{x} + 3 = 8

Problem 15

2x=62\sqrt{x} = 6

Problem 16

x−4=0\sqrt{x - 4} = 0

Problem 17

Square area 81; side?

Problem 18

v = sqrt(2*10*5)?

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 19

3x+1=4\sqrt{3x + 1} = 4

Problem 20

x−2=6\sqrt{x - 2} = 6

Problem 21

5x−1=7\sqrt{5x - 1} = 7

Problem 22

x+10+1=6\sqrt{x + 10} + 1 = 6

Problem 23

2x−3=5\sqrt{2x - 3} = 5

Problem 24

x=x−2(check for extraneous solutions)\begin{gathered}\sqrt{x} = x - 2 \\ \text{(check for extraneous solutions)}\end{gathered}

Problem 25

x−3=x−5(check for extraneous solutions)\begin{gathered}\sqrt{x - 3} = x - 5 \\ \text{(check for extraneous solutions)}\end{gathered}

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