Further Mathematics Questions and Answers

\(2x^{2} - 5x + m = 0\)

\(a = 2, b = -5, c = m\)

\(\alpha + \beta = \frac{-b}{a} = \frac{5}{2}\)

\(\alpha \beta = \frac{c}{a} = \frac{m}{2}\)

\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta\)

= \((\frac{5}{2})^{2} - 2(\frac{m}{2}) \)

= \(\frac{25}{4} - m\)

Rationalizing \(\frac{2}{3 - \sqrt{7}}\) by multiplying through with \(3 + \sqrt{7}\),

\(\frac{2}{3 - \sqrt{7}} \frac{(3 + \sqrt{7})}{(3 + \sqrt{7})} = \frac{6 + 2\sqrt{7}}{9 - 7}\)

= \(\frac{6 + 2\sqrt{7}}{2} = 3 + \sqrt{7}\)

All other options given are commutative i.e. \(a * b = b * a\), except option D.

\(a * b = a - b + ab\)

\(b * a = b - a + ba\)

\(a - b = -(b - a) \neq b - a\)

\((2 + x)^{6} \)

\(x^{4} = ^{6}C_{2}(2^{2})(x^{4}) = 15 \times 4 = 60\)

\(\sin \theta = \frac{\sqrt{3}}{2} \implies opp = \sqrt{3}; hyp = 2\)

\(adj^{2} = 2^{2} - (\sqrt{3})^{2} = 1 \implies adj = 1\)

\(\cos \theta = \frac{1}{2}\)

\(\sin 2\theta = \sin (180 - \theta) = \sin \theta = \frac{\sqrt{3}}{2}\)

\(\cos 2\theta = \cos (180 - \theta) = -\cos \theta = -\frac{1}{2}\)

\(\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\)

= \(- \sqrt{3}\)