Radical Equations Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
Solve 2x+1=xโˆ’1\sqrt{2x + 1} = x - 1.

Solution

  1. 1
    Step 1: Square both sides: 2x+1=(xโˆ’1)2=x2โˆ’2x+12x + 1 = (x-1)^2 = x^2 - 2x + 1.
  2. 2
    Step 2: Rearrange: 0=x2โˆ’4x0 = x^2 - 4x, so x(xโˆ’4)=0x(x - 4) = 0. Solutions: x=0x = 0 or x=4x = 4.
  3. 3
    Step 3: Check x=0x = 0: 1=1\sqrt{1} = 1 but 0โˆ’1=โˆ’10 - 1 = -1. 1โ‰ โˆ’11 \neq -1. Extraneous! โœ—
  4. 4
    Step 4: Check x=4x = 4: 9=3\sqrt{9} = 3 and 4โˆ’1=34 - 1 = 3. โœ“

Answer

x=4x = 4
Squaring both sides can introduce extraneous solutions. Here x=0x = 0 satisfies the squared equation but not the original, because the right side was negative. Always verify each solution.

About Radical Equations

Equations with a variable under a radical sign, solved by isolating the radical, squaring both sides, and checking for extraneous solutions.

Learn more about Radical Equations โ†’

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