Radicals and Rational Expressions

Solving Radical Equations

Solve √(x+2)=x. Square both sides: x+2=x², rearrange to x²-x-2=0, roots x=2 and x=-1. Check each in the ORIGINAL: x=2 gives √4=2 (valid); x=-1 gives √1=1≠-1 (extraneous).

solve-radical-equation Squaring can introduce extraneous solutions. Always verify every candidate root in the ORIGINAL equation, not the squared form.

Example

Square both sides, solve, then check for extraneous roots.

highlighted = computed this step

Step 1 — Set up

Set up the expression.

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

Step 2 — Square both sides

Square both sides to remove the radical.

x+2=x2x+ \hl{2} = \hlmath{x^{2}}

Step 3 — Rearrange

Move all terms to one side: subtract x and 2 to get 0.

x2−x−2=0x^{2} -x- \hl{2} = \hl{0}

Step 4 — Candidate roots

Solve the quadratic: candidates are 2 and negative 1.

x=2orx=−1x= \hl{2} \quad \text{or}\quad x=- \hl{1}

Step 5 — Check valid root

Check x = 2: square root of 4 is 2, so it works.

4=2=2\sqrt{ 4 }= \hl{2} = \hl{2}

Step 6 — Check extraneous root

Check x = negative 1: square root of 1 is 1, not negative 1.

1=1≠−1\sqrt{ 1 }= \hl{1} \ne - \hl{1}

Step 7 — Solution

Keep x = 2 and reject the extraneous root.

x=2valid; reject extraneous rootx= \hl{2} \quad \hl{valid; reject extraneous root}