(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.
Explanation
(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)