(a) If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} + 5x - 6 = 0\), find the equation whose roots are \((\alpha - 2)\) and \((\beta - 2)\).
(b) Given that \(\int_{0} ^{k} (x^{2} - 2x) \mathrm {d} x = 4\), find the values of k.
(a) \(2x^{2} + 5x - 6 = 0\)
\(a = 2 ; b = 5 ; c = -6\)
\(\alpha \beta = \frac{c}{a} = \frac{-6}{2} = -3\)
\(\alpha + \beta = \frac{-b}{a} = \frac{-5}{2}\)
Product of new roots : \((\alpha - 2)(\beta - 2) = \alpha \beta - 2(\alpha + \beta) + 4\)
= \(-3 - 2(-\frac{5}{2}) + 4 = 6\)
Sum of new roots : \((\alpha - 2) + (\beta - 2) = \alpha + \beta - 4\)
= \(\frac{-5}{2} - 4 = -\frac{13}{2}\)
Equation : \(x^{2} - (sum) x + product = 0\)
\(x^{2} - (-\frac{13}{2} x) + 6 = 0\)
\(x^{2} + \frac{13}{2} x + 6 = 0 \implies 2x^{2} + 13x + 12 = 0\)
(b) \(\int_{0} ^{k} (x^{2} - 2x) \mathrm {d} x = \frac{4}{3}\)
\([\frac{x^{3}}{3} - x^{2}]_{0} ^{k} = \frac{4}{3}\)
\(\frac{k^{3}}{3} - k^{2} = \frac{4}{3}\)
\(k^{3} - 3k^{2} = 4 \implies k^{3} - 3k^{2} - 4 = 0\)
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