WAEC 2009 · Paper 2 · Q1✱✱

(a)
Solve $2x^2 + x - 6 < 0$ .
Show the answer
$-2 < x < \frac32$
(b)
Express $\dfrac{5 - 2\sqrt{10}}{3\sqrt5 + \sqrt2}$ in the form $m\sqrt2 + n\sqrt5$ , where $m$ and $n$ are rational numbers.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Factorise the quadratic: $2x^2 + x - 6 = (2x - 3)(x + 2)$ .
The expression is zero at $x = \frac32$ and $x = -2$ .
These split the number line into three intervals.
A product is negative when its two brackets have opposite signs.
That happens only between the roots: for example, $x = 0$ gives $(-3)(2) = -6 < 0$ .
So the solution is $-2 < x < \frac32$ .

(b)

Multiply the top and bottom by the conjugate of the denominator, $3\sqrt5 - \sqrt2$ .
Bottom: $(3\sqrt5)^2 - (\sqrt2)^2 = 45 - 2 = 43$ .
Top: $(5 - 2\sqrt{10})(3\sqrt5 - \sqrt2) = 15\sqrt5 - 5\sqrt2 - 6\sqrt{50} + 2\sqrt{20}$ .
Simplify the surds: $6\sqrt{50} = 30\sqrt2$ and $2\sqrt{20} = 4\sqrt5$ , so the top is $19\sqrt5 - 35\sqrt2$ .
So the fraction is $-\frac{35}{43}\sqrt2 + \frac{19}{43}\sqrt5$ : $m = -\frac{35}{43}$ and $n = \frac{19}{43}$ .

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