(a) Solve : \(2x^{2} + x - 6 < 0\) (b) Express \(\frac{5 - 2\sqrt{10}}{3\sqrt{5}

FURTHER MATHEMATICS
WAEC 2009

(a) Solve : \(2x^{2} + x - 6 < 0\)

(b) Express \(\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}}\) in the form \(m\sqrt{2} + n\sqrt{5}\) where m and n are rational numbers.

Explanation

(a) \(2x^{2} + x - 6 < 0 \implies 2x^{2} + 4x - 3x - 6 < 0\)

\(2x(x + 2) - 3(x + 2) < 0 \implies (2x - 3)(x + 2) < 0\)

If \((2x - 3)(x + 2) < 0\), then either (2x - 3) < 0 or (x + 2) < 0.

\((2x - 3) < 0 \implies x < \frac{3}{2}\), then \((x + 2) > 0 \implies x > -2\)

\((x + 2) < 0 \implies x < -2\), then \((2x - 3) > 0 \implies x > \frac{3}{2}\)

Check: Let x = -3,

\((2x - 3)(x + 2) = (-9)(-1) = 9 > 0\)

Let x = 1,

\((2x - 3)(x + 2) = (-1)(3) = -3 < 0\)

\(\therefore -2 < x < \frac{3}{2}\).

(b) \(\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}}\)

= \((\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}})(\frac{3\sqrt{5} - \sqrt{2}}{3\sqrt{5} - \sqrt{2}})\)

= \(\frac{15\sqrt{5} - 5\sqrt{2} - 30\sqrt{2} + 4\sqrt{5}}{9(5) - 3\sqrt{10} + 3\sqrt{10} - 2}\)

= \(\frac{19\sqrt{5} - 35\sqrt{2}}{45 - 2}\)

= \(\frac{19\sqrt{5} - 35\sqrt{2}}{43}\)

= \(\frac{-35}{43} \sqrt{2} + \frac{19}{43} \sqrt{5}\)

\(\therefore m = -\frac{-35}{43} ; n = \frac{19}{43}\).



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