Q13. If the zeroes of the polynomial \(ax^2 + bx +
\frac{2a}{b}\) are reciprocal of each other, then the value of \(b\) is
(A) 2
(B) \(\frac{1}{2}\)
(C) -2
(D) \(-\frac{1}{2}\)
Q31. Find the zeroes of the polynomial \(r(x) = 4x^2 + 3x -
1\). Hence, write a polynomial whose zeroes are reciprocal of the zeroes of polynomial \(r(x)\).
\(4x^2 + 3x - 1 = (4x-1)(x+1) = 0\). Zeroes are \(x = \frac{1}{4}\) and \(x = -1\).
New zeroes (reciprocals): 4 and -1. Sum = 3, Product = -4.
Required Polynomial = \(x^2 - 3x - 4\).