If the quadratic equation \((2x - 1) - p(x^{2} + 2) = 0\), where p is a constant, has real roots :
(a) show that \(2p^{2} + p - 1 < 0\);
(b) find the values of p.
Show Answer Show Explanation Explanation (a) \((2x - 1) - p(x^{2} + 2) = 0\) \(2x - 1 - px^{2} - 2p = 0\) \(px^{2} - 2x + (2p + 1) = 0\) For real roots, \(b^{2} - 4ac \geq 0\) \(2^{2} - 4(p)(2p + 1) \geq 0\) \(4 - 8p^{2} - 4p \geq 0 \implies 2p^{2} + p - 1 \leq 0\) (b) \(2p^{2} - p + 2p - 1 \leq 0 \implies p(2p - 1) + 1(2p - 1) \leq 0\) \((p + 1)(2p - 1) \leq 0 \) \(2p - 1 \leq 0 \implies p \leq \frac{1}{2}\) \(p + 1 \leq 0 \implies p \leq -1\) Check : For \(p \leq 1 , \text{let p = -2}\) \(2(-2)^{2} + (-2) - 1 = 8 - 2 - 1 = 5 \) For \(p \leq \frac{1}{2}, \text{let p = 0}\) \(2(0^{2}) + 0 - 1 = -1 \leq 0\) \(\therefore -1 \leq p \leq \frac{1}{2}\).