WAEC 2012 · Paper 2 · Q10✱✱

(a)
Simplify $\dfrac{4\sqrt7 + 3\sqrt2}{10 + 2\sqrt{14}}$ .
Show the answer
$\dfrac{14\sqrt7 - 13\sqrt2}{22}$
(b)
Given that $y = \dfrac{px^2 + q}{x^4}$ , where $p$ and $q$ are constants, show that $x^2\dfrac{d^2y}{dx^2} + 7x\dfrac{dy}{dx} + 8y = 0$ .
Model answer
Write $y = px^{-2} + qx^{-4}$ . Then $\dfrac{dy}{dx} = -2px^{-3} - 4qx^{-5}$ and $\dfrac{d^2y}{dx^2} = 6px^{-4} + 20qx^{-6}$ .

$x^2\dfrac{d^2y}{dx^2} = 6px^{-2} + 20qx^{-4}$ , $\quad 7x\dfrac{dy}{dx} = -14px^{-2} - 28qx^{-4}$ , $\quad 8y = 8px^{-2} + 8qx^{-4}$ .

Adding: $(6 - 14 + 8)px^{-2} + (20 - 28 + 8)qx^{-4} = 0$ , as required.

Worked solution (try it first)

Multiply the top and the bottom by the conjugate of the bottom, $10 - 2\sqrt{14}$ .
The bottom: $10^2 - (2\sqrt{14})^2 = 100 - 56 = 44$ .
The top: $(4\sqrt7 + 3\sqrt2)(10 - 2\sqrt{14}) = 40\sqrt7 - 8\sqrt{98} + 30\sqrt2 - 6\sqrt{28}$ .
Simplify the surds: $\sqrt{98} = 7\sqrt2$ and $\sqrt{28} = 2\sqrt7$ , so the top is $40\sqrt7 - 56\sqrt2 + 30\sqrt2 - 12\sqrt7 = 28\sqrt7 - 26\sqrt2$ .
Divide by 44 and simplify: $\dfrac{28\sqrt7 - 26\sqrt2}{44} = \dfrac{14\sqrt7 - 13\sqrt2}{22}$ .

Write $y$ as powers of $x$ : $y = px^{-2} + qx^{-4}$ .
Differentiate: $\dfrac{dy}{dx} = -2px^{-3} - 4qx^{-5}$ , and again: $\dfrac{d^2y}{dx^2} = 6px^{-4} + 20qx^{-6}$ .
Multiply: $x^2\dfrac{d^2y}{dx^2} = 6px^{-2} + 20qx^{-4}$ , $7x\dfrac{dy}{dx} = -14px^{-2} - 28qx^{-4}$ and $8y = 8px^{-2} + 8qx^{-4}$ .
Add the three: the $px^{-2}$ terms give $6 - 14 + 8 = 0$ and the $qx^{-4}$ terms give $20 - 28 + 8 = 0$ .
So $x^2\dfrac{d^2y}{dx^2} + 7x\dfrac{dy}{dx} + 8y = 0$ , as required.