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.
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}).