Subject that is squared

If the new subject is squared, isolate the square and take the square root of both sides. A positive number has two square roots, so \(x^2=A\) gives \(x=\pm\sqrt{A}\), for \(A\ge0\). Keep both signs unless the context rules one out.

Why: \(3^2=9\) and \((-3)^2=9\), so both \(3\) and \(-3\) satisfy \(x^2=9\). A length, a speed or a mass cannot be negative, so in those contexts only the positive root is kept.

Method

1. Rearrange so the square stands alone on one side.
2. Take the square root of both sides and write \(\pm\).
3. Use the context to decide whether the negative root is possible.
4. State the condition that the expression under the root is not negative.

Example

Make \(x\) the subject of \(y=3x^2-5\).

\(\begin{aligned}y+5&=3x^2 \\ x^2&=\dfrac{y+5}{3} \\ x&=\pm\sqrt{\dfrac{y+5}{3}},\qquad y\ge-5\end{aligned}\)

In context: a square photo of side \(s\) cm with a \(2\) cm border all round has total area \(A=(s+4)^2\). Taking the positive root, \(s+4=\sqrt A\), so \(s=\sqrt A-4\).

Check: forgetting \(\pm\) loses half the answers. Also \(\sqrt{a^2+b^2}\) is not \(a+b\): with \(a=b=1\), \(\sqrt{2}\) is not \(2\).