Equations and inequalities

Root equations

Root equations need one extra step that equations without roots do not: squaring can create solutions that are not real solutions, so you must check every result.

The rules

• Isolate the root on one side first.
• Square both sides: (√a)² = a.
• Solve the resulting equation.
• Check every solution in the original equation and drop any that do not work.

Practice questions with solutions

Generated by the same engine as the Kollego course and solvable without a calculator. Pick an answer in your head, then open the solution.

1. 1. Solve for x: √(3x − 5) = 2

   • a2
   • b4
   • c-3
   • d3
   Show solution

   Answer: 3

   Square both sides to remove the root, then isolate x. Since the right side is already ≥ 0 here, squaring introduces no extraneous root, but always check that the expression under the root is ≥ 0 for your solution.

   Square both sides: 3x − 5 = 2² = 4. x = (4 + 5) / 3 = 3. Check: 3(3) − 5 = 4 = 2², and √4 = 2 ✓.

2. 2. Solve for x: √(2x + 29) = 5

   • a-1
   • b-2
   • c-3
   • d2
   Show solution

   Answer: -2

   Square both sides to remove the root, then isolate x. Since the right side is already ≥ 0 here, squaring introduces no extraneous root, but always check that the expression under the root is ≥ 0 for your solution.

   Square both sides: 2x + 29 = 5² = 25. x = (25 − 29) / 2 = -2. Check: 2(-2) + 29 = 25 = 5², and √25 = 5 ✓.

3. 3. Solve for x: √(3x + 9) = 3

   • a2
   • b1
   • c0
   • d-1
   Show solution

   Answer: 0

   Square both sides to remove the root, then isolate x. Since the right side is already ≥ 0 here, squaring introduces no extraneous root, but always check that the expression under the root is ≥ 0 for your solution.

   Square both sides: 3x + 9 = 3² = 9. x = (9 − 9) / 3 = 0. Check: 3(0) + 9 = 9 = 3², and √9 = 3 ✓.

4. 4. Solve for x: √(2x + 18) = 4

   • a0
   • b-2
   • c-1
   • d1
   Show solution

   Answer: -1

   Square both sides to remove the root, then isolate x. Since the right side is already ≥ 0 here, squaring introduces no extraneous root, but always check that the expression under the root is ≥ 0 for your solution.

   Square both sides: 2x + 18 = 4² = 16. x = (16 − 18) / 2 = -1. Check: 2(-1) + 18 = 16 = 4², and √16 = 4 ✓.

5. 5. Solve for x: √(2x + 44) = 6

   • a-4
   • b-3
   • c4
   • d-5
   Show solution

   Answer: -4

   Square both sides to remove the root, then isolate x. Since the right side is already ≥ 0 here, squaring introduces no extraneous root, but always check that the expression under the root is ≥ 0 for your solution.

   Square both sides: 2x + 44 = 6² = 36. x = (36 − 44) / 2 = -4. Check: 2(-4) + 44 = 36 = 6², and √36 = 6 ✓.

6. 6. Solve for x: √(2x + 34) = 6

   • a-1
   • b1
   • c0
   • d2
   Show solution

   Answer: 1

   Square both sides to remove the root, then isolate x. Since the right side is already ≥ 0 here, squaring introduces no extraneous root, but always check that the expression under the root is ≥ 0 for your solution.

   Square both sides: 2x + 34 = 6² = 36. x = (36 − 34) / 2 = 1. Check: 2(1) + 34 = 36 = 6², and √36 = 6 ✓.

Common mistakes

• Squaring before isolating the root.
• Squaring a sum as if it were a product: (a + b)² needs the binomial formula.
• Skipping the check and keeping a false solution.

Related topics

All maths topics · T-Kurs · M-Kurs · W-Kurs