Equations with the unknown under a square root

When the unknown is under a square root, make the root the subject, square both sides and solve the equation that results. Squaring can add values that do not satisfy the original equation, so every candidate is checked by substituting it into the original equation.

Why extra roots appear: squaring cannot be undone. If \(A=B\) then \(A^2=B^2\), but \(A^2=B^2\) only gives \(A=B\) or \(A=-B\). So the squared equation also holds the solutions of the equation with the other side negated. A square root is never negative, so a candidate that makes the other side negative must be rejected.

Method

1. Make the square root the subject: write the equation as \(\sqrt{\text{expression}}=\text{the rest}\).
2. Square both sides, squaring the whole of each side.
3. Collect the terms into a quadratic equation and solve it.
4. Substitute each candidate into the original equation. Keep it if both sides are equal; reject it, with the reason, if they are not.
5. State the solution.

Example

Solve \(\sqrt{5x+6}=2x\).

The square root is already the subject. Square both sides:

\(\begin{aligned} 5x+6&=4x^2 \\ 4x^2-5x-6&=0 \\ (4x+3)(x-2)&=0 \end{aligned}\)

The candidates are \(x=2\) and \(x=-\tfrac34\).

Check \(x=2\): \(\sqrt{16}=4\) and \(2(2)=4\). Both sides are equal, so \(x=2\) is a solution.

Check \(x=-\tfrac34\): \(\sqrt{\tfrac94}=\tfrac32\) but \(2\left(-\tfrac34\right)=-\tfrac32\). A square root cannot be negative, so \(x=-\tfrac34\) is rejected.

\(\therefore\ x=2\)

Showing that a claim is false

A claim made "for every \(x\)" is false as soon as one value fails. To show this, give that value (a counterexample), work out both sides, and state that they are not equal. Values for which the claim happens to hold never prove it for every \(x\).

For example, the claim "if \(a^2=b^2\), then \(a=b\)" is false: take \(a=3\) and \(b=-3\). Then \(a^2=9=b^2\), but \(a\ne b\). This is exactly why a squared equation needs checking.