Solving Radical Equations

Equations containing radicals

Solving Radical Equations

Strategy

  1. Isolate the radical on one side
  2. Raise both sides to the appropriate power
  3. Solve the resulting equation
  4. Check for extraneous solutions

Why Check?

Squaring both sides can introduce extraneous solutions.

Always substitute back into the original equation!

Multiple Radicals

If there are two radicals:

  1. Isolate one radical
  2. Square both sides
  3. Isolate the remaining radical
  4. Square again
  5. Solve and check

Domain Restrictions

For x\sqrt{x}: must have xโ‰ฅ0x \geq 0

For x3\sqrt[3]{x}: xx can be any real number

Example

Solve: 2x+3=5\sqrt{2x + 3} = 5

Square both sides: 2x+3=252x + 3 = 25 2x=222x = 22 x=11x = 11

Check: 2(11)+3=25=5\sqrt{2(11) + 3} = \sqrt{25} = 5 โœ“

๐Ÿ“š Practice Problems

1Problem 1easy

โ“ Question:

Solve: โˆšx = 5

๐Ÿ’ก Show Solution

Step 1: Square both sides: (โˆšx)ยฒ = 5ยฒ x = 25

Step 2: Check the solution: โˆš25 = 5 โœ“

Answer: x = 25

2Problem 2easy

โ“ Question:

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

๐Ÿ’ก Show Solution

Square both sides: x+5=16x + 5 = 16

Solve for xx: x=11x = 11

Check: 11+5=16=4\sqrt{11 + 5} = \sqrt{16} = 4 โœ“

Answer: x=11x = 11

3Problem 3easy

โ“ Question:

Solve: โˆš(x + 3) = 4

๐Ÿ’ก Show Solution

Step 1: Square both sides: (โˆš(x + 3))ยฒ = 4ยฒ x + 3 = 16

Step 2: Solve for x: x = 16 - 3 x = 13

Step 3: Check the solution: โˆš(13 + 3) = โˆš16 = 4 โœ“

Answer: x = 13

4Problem 4medium

โ“ Question:

Solve: 3xโˆ’2+4=10\sqrt{3x - 2} + 4 = 10

๐Ÿ’ก Show Solution

Step 1: Isolate the radical 3xโˆ’2=6\sqrt{3x - 2} = 6

Step 2: Square both sides 3xโˆ’2=363x - 2 = 36

Step 3: Solve 3x=383x = 38 x=383x = \frac{38}{3}

Check: 3(383)โˆ’2+4=38โˆ’2+4=36+4=6+4=10\sqrt{3(\frac{38}{3}) - 2} + 4 = \sqrt{38 - 2} + 4 = \sqrt{36} + 4 = 6 + 4 = 10 โœ“

Answer: x=383x = \frac{38}{3}

5Problem 5medium

โ“ Question:

Solve: โˆš(2x - 1) = x - 2

๐Ÿ’ก Show Solution

Step 1: Square both sides: (โˆš(2x - 1))ยฒ = (x - 2)ยฒ 2x - 1 = xยฒ - 4x + 4

Step 2: Rearrange to standard form: 0 = xยฒ - 4x - 2x + 4 + 1 0 = xยฒ - 6x + 5

Step 3: Factor: 0 = (x - 5)(x - 1)

Step 4: Solve: x = 5 or x = 1

Step 5: Check x = 5: โˆš(2(5) - 1) = โˆš9 = 3 5 - 2 = 3 โœ“

Step 6: Check x = 1: โˆš(2(1) - 1) = โˆš1 = 1 1 - 2 = -1 โœ— x = 1 is EXTRANEOUS

Answer: x = 5

6Problem 6medium

โ“ Question:

Solve: โˆš(x + 7) - โˆš(x - 5) = 2

๐Ÿ’ก Show Solution

Step 1: Isolate one radical: โˆš(x + 7) = 2 + โˆš(x - 5)

Step 2: Square both sides: x + 7 = (2 + โˆš(x - 5))ยฒ x + 7 = 4 + 4โˆš(x - 5) + (x - 5) x + 7 = x - 1 + 4โˆš(x - 5)

Step 3: Simplify: x + 7 - x + 1 = 4โˆš(x - 5) 8 = 4โˆš(x - 5) 2 = โˆš(x - 5)

Step 4: Square again: 4 = x - 5 x = 9

Step 5: Check x = 9: โˆš(9 + 7) - โˆš(9 - 5) = โˆš16 - โˆš4 = 4 - 2 = 2 โœ“

Answer: x = 9

7Problem 7hard

โ“ Question:

Solve: x+7=xโˆ’5\sqrt{x + 7} = x - 5

๐Ÿ’ก Show Solution

Step 1: Square both sides x+7=(xโˆ’5)2x + 7 = (x - 5)^2 x+7=x2โˆ’10x+25x + 7 = x^2 - 10x + 25

Step 2: Rearrange to standard form 0=x2โˆ’11x+180 = x^2 - 11x + 18

Step 3: Factor 0=(xโˆ’9)(xโˆ’2)0 = (x - 9)(x - 2)

Step 4: Solve x=9ย orย x=2x = 9 \text{ or } x = 2

Step 5: Check both solutions

For x=9x = 9: 9+7=16=4\sqrt{9 + 7} = \sqrt{16} = 4 and 9โˆ’5=49 - 5 = 4 โœ“

For x=2x = 2: 2+7=9=3\sqrt{2 + 7} = \sqrt{9} = 3 and 2โˆ’5=โˆ’32 - 5 = -3 โœ—

Answer: x=9x = 9 only (x = 2 is extraneous)

8Problem 8hard

โ“ Question:

Solve: โˆ›(x + 1) = 2

๐Ÿ’ก Show Solution

Step 1: Cube both sides: (โˆ›(x + 1))ยณ = 2ยณ x + 1 = 8

Step 2: Solve for x: x = 7

Step 3: Check: โˆ›(7 + 1) = โˆ›8 = 2 โœ“

Step 4: Note about odd roots: Unlike even roots, odd roots can have negative radicands For example, โˆ›(-8) = -2

Answer: x = 7