Subject under a square root or cube root

When the new subject is under a root, first isolate the root on one side, then square both sides (for a square root) or cube both sides (for a cube root). After that, rearrange as usual.

Why: squaring undoes a square root, \(\left(\sqrt{A}\right)^2=A\), and cubing undoes a cube root, \(\left(\sqrt[3]{A}\right)^3=A\). Squaring a side that still has another term does not remove the root: \(\left(\sqrt{a}+3\right)^2\) is not \(a+9\).

Method

1. Move every term outside the root to the other side, and clear any number multiplying the root.
2. Square (or cube) both sides. Square the whole of the other side, using brackets.
3. Rearrange the result: clear fractions, collect the subject and factorise if it appears more than once.
4. Note any condition: a square root is never negative.

Example

Make \(x\) the subject of \(y=\dfrac{3}{\sqrt{x-1}}\).

\(\begin{aligned}y\sqrt{x-1}&=3 \\ \sqrt{x-1}&=\dfrac3y \\ x-1&=\dfrac{9}{y^2} \\ x&=\dfrac{9}{y^2}+1,\qquad y>0\end{aligned}\)

Cube root: the radius of a sphere of volume \(V\) is \(r=\sqrt[3]{\dfrac{3V}{4\pi}}\). Cubing gives \(r^3=\dfrac{3V}{4\pi}\), so \(V=\dfrac{4\pi r^3}{3}\).

Check: square the whole side: \(\left(\dfrac3y\right)^2=\dfrac{9}{y^2}\), not \(\dfrac{3}{y^2}\). Likewise \((p-3)^2\) is \(p^2-6p+9\), not \(p^2-9\). The condition \(y>0\) comes from \(\dfrac{3}{\sqrt{x-1}}>0\).