(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 ratio...
Question 1 Report
(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.
Answer Details
(a) Factorise \(2x^2 + x - 6 = (2x - 3)(x + 2)\). Roots at \(x = \dfrac{3}{2}\) and \(x = -2\). Since the parabola opens upward, it is negative between the roots:
\[-2 < x < \frac{3}{2}\](b) Multiply numerator and denominator by the conjugate \((3\sqrt5 - \sqrt2)\):
\[\text{Denominator} = (3\sqrt5)^2 - (\sqrt2)^2 = 45 - 2 = 43\] \[\text{Numerator} = (5 - 2\sqrt{10})(3\sqrt5 - \sqrt2) = 15\sqrt5 - 5\sqrt2 - 6\sqrt{50} + 2\sqrt{20}\]Since \(\sqrt{50} = 5\sqrt2\) and \(\sqrt{20} = 2\sqrt5\):
\[= 15\sqrt5 - 5\sqrt2 - 30\sqrt2 + 4\sqrt5 = 19\sqrt5 - 35\sqrt2\]Therefore
\[\frac{5 - 2\sqrt{10}}{3\sqrt5 + \sqrt2} = \frac{19\sqrt5 - 35\sqrt2}{43} = -\frac{35}{43}\sqrt2 + \frac{19}{43}\sqrt5\]So \(m = -\dfrac{35}{43}\) and \(n = \dfrac{19}{43}\).
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