Quadratics: Question 10

Worked solution

Part (a): Making the substitution

Let $u = \sqrt{x}$, so $x \ge 0$ and $u \ge 0$. Then $x = u^2$, so the equation $6x - 5\sqrt{x} - 6 = 0$ becomes:

$6u^2 - 5u - 6 = 0$

as required.

Part (b): Solving the quadratic in $u$

Factorise, looking for two numbers with product $6\times(-6)=-36$ and sum $-5$: these are $-9$ and $4$.

$6u^2 - 9u + 4u - 6 = 0$

$3u(2u-3) + 2(2u-3) = 0$

$(2u-3)(3u+2) = 0$

$u = \frac{3}{2} \qquad \text{or} \qquad u = -\frac{2}{3}$

Check using the quadratic formula: with $a=6$, $b=-5$, $c=-6$,

$\text{discriminant} = (-5)^2 - 4(6)(-6) = 25 + 144 = 169$

$u = \frac{5 \pm \sqrt{169}}{12} = \frac{5 \pm 13}{12} = \frac{18}{12} \text{ or } \frac{-8}{12} = \frac{3}{2} \text{ or } -\frac{2}{3}$

This agrees exactly with the factorised solution.

Part (c): Returning to $x$, and rejecting an extraneous branch

Recall $u = \sqrt{x}$. By definition, the (principal) square root is never negative, so $u \ge 0$ for every valid $x$.

• $u = -\dfrac{2}{3}$ would require $\sqrt{x} = -\dfrac{2}{3}$, which is impossible, so this value of $u$ must be rejected.
• $u = \dfrac{3}{2}$ gives $\sqrt{x} = \dfrac{3}{2}$, so $x = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$.

So the only real solution of $6x - 5\sqrt{x} - 6 = 0$ (with $x\ge0$) is $x = \dfrac{9}{4}$.

Check: at $x=\tfrac94=2.25$, $\sqrt{x}=1.5$, so $6x-5\sqrt{x}-6 = 6(2.25)-5(1.5)-6 = 13.5-7.5-6=0$. ✓

Final answers

• (a) $6u^2 - 5u - 6 = 0$ (shown)
• (b) $u = \boxed{\dfrac{3}{2}}$ or $u = \boxed{-\dfrac{2}{3}}$
• (c) $x = \boxed{\dfrac{9}{4}}$ (rejecting $u=-\tfrac23$ as $\sqrt{x}$ cannot be negative)