1. **State the problem:** Given the equation $(2x - 1) - p(x^2 + 2) = 0$, where $p$ is a constant, we need to show that $2p^2 + p - 1 \leq 0$ for the equation to have real roots, and then find the value of $p$. 2. **Rewrite the equation:** Expand and rearrange the equation into standard quadratic form in $x$: $$ (2x - 1) - p(x^2 + 2) = 0 \implies 2x - 1 - p x^2 - 2p = 0 $$ Rearranged: $$ -p x^2 + 2x - (1 + 2p) = 0 $$ Multiply both sides by $-1$ to get: $$ p x^2 - 2x + (1 + 2p) = 0 $$ 3. **Identify coefficients:** The quadratic in $x$ is: $$ p x^2 - 2x + (1 + 2p) = 0 $$ where $a = p$, $b = -2$, and $c = 1 + 2p$. 4. **Condition for real roots:** For the quadratic to have real roots, the discriminant must be non-negative: $$ \Delta = b^2 - 4ac \geq 0 $$ Substitute values: $$ (-2)^2 - 4 \times p \times (1 + 2p) \geq 0 $$ $$ 4 - 4p - 8p^2 \geq 0 $$ Divide both sides by 4: $$ 1 - p - 2p^2 \geq 0 $$ Rewrite: $$ -2p^2 - p + 1 \geq 0 $$ Multiply both sides by $-1$ (reversing inequality): $$ 2p^2 + p - 1 \leq 0 $$ This proves part (a). 5. **Find values of $p$:** Solve the quadratic inequality: $$ 2p^2 + p - 1 = 0 $$ Use quadratic formula: $$ p = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} $$ Two roots: $$ p_1 = \frac{-1 + 3}{4} = \frac{2}{4} = 0.5 $$ $$ p_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 $$ 6. **Interpret inequality:** Since $2p^2 + p - 1$ is a parabola opening upwards, the inequality $2p^2 + p - 1 \leq 0$ holds between the roots: $$ -1 \leq p \leq 0.5 $$ **Final answers:** - (a) $2p^2 + p - 1 \leq 0$ for real roots. - (b) $p$ lies in the interval $[-1, 0.5]$.