Question

Find the value of p and q. (a) x²+ px - 2 = (x - 1)(x +q)​

Answer 1

Answer:

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Answer 2

$\textsf{\large{\underline{Solution}:}}$

Given:

$\rm \longrightarrow {x}^{2} + px - 2 = (x - 1)(x + q)$

$\rm \longrightarrow {x}^{2} + px - 2 = x(x + q) - 1(x + q)$

$\rm \longrightarrow {x}^{2} + px - 2 = {x}^{2} + qx- x - q$

$\rm \longrightarrow px - 2 = qx- x - q$

$\rm \longrightarrow (p)x + ( - 2)= (q - 1)x + ( - q)$

Comparing both sides:

$\rm \longrightarrow \begin{cases} \rm p = q - 1 \\ \rm - q = - 2 \end{cases}$

$\rm \longrightarrow q = 2$

$\rm \longrightarrow p = q - 1$

$\rm \longrightarrow p = 1$

Therefore:

$\rm \longrightarrow (p, q) = (1,2)$

$\textsf{\large{\underline{Verification}:}}$

Put p = 1 in LHS, we get:

$\rm = {x}^{2} + x - 2$

Put q = 2 in RHS, we get:

$\rm = (x - 1)(x + 2)$

$\rm = {x}^{2} + 2x - x - 2$

$\rm = {x}^{2} + x - 2$

Therefore, our answer is correct (Verified)