Solving Radical Equations

Strategy: isolate the radical, square both sides, CHECK answers

Squaring can introduce extraneous solutions (false answers), so checking is mandatory — not optional.

Step-by-step: Solve $\sqrt{2x + 3} = x - 1$

Step 1 — The radical is already isolated on the left side. Good.

Step 2 — Square both sides to eliminate the square root:

$(\sqrt{2x+3})^2 = (x-1)^2$ $2x + 3 = x^2 - 2x + 1$

Step 3 — Rearrange to standard form:

$0 = x^2 - 4x - 2$

Step 4 — Solve with the quadratic formula: $a=1, b=-4, c=-2$.

$x = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}$

$x_1 = 2 + \sqrt{6} \approx 4.449, \quad x_2 = 2 - \sqrt{6} \approx -0.449$

Step 5 — CHECK both answers in the original equation. This step is critical!

Check $x_1 = 2 + \sqrt{6} \approx 4.449$:

LHS: $\sqrt{2(4.449) + 3} = \sqrt{11.899} \approx 3.449$

RHS: $4.449 - 1 = 3.449$ ✓ Equal!

Check $x_2 = 2 - \sqrt{6} \approx -0.449$:

LHS: $\sqrt{2(-0.449) + 3} = \sqrt{2.102} \approx 1.449$

RHS: $-0.449 - 1 = -1.449$ ✗ Not equal! ($1.449 \neq -1.449$)

$x_2$ is extraneous — it was created by the squaring step. Reject it!

Final answer: $x = 2 + \sqrt{6}$ only.

Why squaring creates extraneous solutions

Squaring is not a reversible operation: $(-3)^2 = 9$ and $3^2 = 9$. When you square both sides, you can't tell if the original side was positive or negative. The check step catches these impostors.

Try it in the visualization

The graphs of $y = \sqrt{2x+3}$ and $y = x-1$ show one intersection (the valid solution) and one non-intersection (where the extraneous root would be). The extraneous root is highlighted in red.