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...
Question 1 Report
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.
Answer Details
(a) Expand the equation \((2x - 1) - p(x^2 + 2) = 0\) into standard quadratic form:
\[-px^2 + 2x - 1 - 2p = 0 \;\Rightarrow\; px^2 - 2x + (1 + 2p) = 0.\]
Here \(a = p,\ b = -2,\ c = 1 + 2p\). For real roots the discriminant must be non-negative:
\[b^2 - 4ac \ge 0 \;\Rightarrow\; (-2)^2 - 4p(1 + 2p) \ge 0 \;\Rightarrow\; 4 - 4p - 8p^2 \ge 0.\]
Divide through by \(4\): \(1 - p - 2p^2 \ge 0\). Multiplying by \(-1\) (reversing the inequality):
\[2p^2 + p - 1 \le 0.\]
(For two distinct real roots the inequality is strict, \(2p^2 + p - 1 < 0\), as stated.)
(b) Solve \(2p^2 + p - 1 = 0\): factorising, \((2p - 1)(p + 1) = 0\), so \(p = \tfrac12\) or \(p = -1\).
Since the coefficient of \(p^2\) is positive, \(2p^2 + p - 1 \le 0\) between the roots:
\[-1 \le p \le \tfrac{1}{2}.\]
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