A Quadratic in Disguise with a Square Root

An equation like $x - 5\sqrt{x} + 6 = 0$ is quadratic in $\sqrt{x}$. Substitute $u = \sqrt{x}$ (so $x = u^2$), solve the quadratic, then square back to $x$.

Solve $x - 5\sqrt{x} + 6 = 0$. [4]

The working

Step 1, substitute $u = \sqrt{x}$, noting $x = (\sqrt{x})^2 = u^2$: $u^2 - 5u + 6 = 0 \quad \text{(M1)}$

Step 2, factorise and solve for $u$: $(u - 2)(u - 3) = 0 \implies u = 2 \ \text{or} \ u = 3 \quad \text{(A1)}$

Both are positive, so both are valid values of $\sqrt{x}$.

Step 3, square back to find $x$ (since $x = u^2$): $x = 2^2 = 4 \qquad \text{or} \qquad x = 3^2 = 9 \quad \text{(A1, A1)}$

Check: $4 - 5\sqrt{4} + 6 = 4 - 10 + 6 = 0$; $\ 9 - 5\sqrt{9} + 6 = 9 - 15 + 6 = 0$. ✓

Where the marks are won and lost

• Recognise $x = (\sqrt{x})^2$, which is what makes the substitution give a clean quadratic.
• Square back: solving gives $\sqrt{x} = 2, 3$, so $x = 4, 9$. Stopping at $u$ loses the final marks.
• Discard any negative $u$, since $\sqrt{x} \ge 0$ (not an issue here, but essential in general).

Common mistakes

• Leaving the answer as $\sqrt{x} = 2, 3$ without squaring.
• Treating $\sqrt{x}$ as if it could be negative and inventing extra roots.
• Trying to square the whole equation first, which creates a messier problem.

Topic home: Quadratic Functions pillar. More: Worked examples.