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.
(a) show that \(2p^{2} + p - 1 < 0\);
(b) find the values of p.
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Correct Answer: Option n
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 \(p \leq \frac{1}{2}, \text{let p = 0}\)
\(2(0^{2}) + 0 - 1 = -1 \leq 0\)
\(\therefore -1 \leq p \leq \frac{1}{2}\).
(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 \(p \leq \frac{1}{2}, \text{let p = 0}\)
\(2(0^{2}) + 0 - 1 = -1 \leq 0\)
\(\therefore -1 \leq p \leq \frac{1}{2}\).