Question

2p² + 5p – 3 = 0, p = 1, $\frac{1}{2}$ , –3.

Answer 1

Given: 2p² + 5p – 3 = 0 and p = 1, $\frac{1}{2}$ , –3.
On substituting p = 1 in L.H.S. of the given equation, we get:
2(1)² + 5(1) – 3
= 2 + 5 – 3
= 4
$\Rightarrow$ L.H.S. $\ne$ R.H.S.
Thus, p = 1 does not satisfy the given equation.
Therefore, p = 1 is not a root of the given quadratic equation.

On substituting p = $\frac{1}{2}$ in L.H.S. of the given equation, we get:
$$\begin{gathered} 2{\left(\frac{1}{2}\right)}^2+5\times \frac{1}{2}-3 \\ =2\times \frac{1}{4}+\frac{5}{2}-3 \\ =\frac{1}{2}+\frac{5}{2}-3 \\ =\frac{1+5-6}{2} \\ =\frac{6-6}{2} \\ =\frac{0}{2} \\ =0 \\ \Rightarrow \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Thus, p = $\frac{1}{2}$ satisfies the given equation.
Therefore, p = $\frac{1}{2}$ is a root of the given quadratic equation.
On substituting p = –3 in L.H.S. of the given equation, we get
2(–3)² + 5(–3) – 3
= 18 – 15 – 3
= 0
$\Rightarrow$ L.H.S. = R.H.S.
Thus, p = –3 satisfies the given equation.
Therefore, p = –3 is a root of the given quadratic equation.