If both x - 2 and x - 1/2 are factors of px² + 5x + r, show that p = r. Class 9

If both x - 2 and x - 1/2 are factors of px² + 5x + r, show that p = r. Class 9

Question 1.

If both x - 2 and x - $\frac{1}{2}$ are factors of px² + 5x + r, show that p = r. Class 9

Solution:

Let q(x) = px² + 5x + r

Given that, x - 2 and x - $\frac{1}{2}$ are the factors of q(x).

∴ By the factor theorem,

q(2) = 0

∴ p(2)² + 5(2) + r = 0

∴ 4p + 10 + r = 0 ...(i)

Also, $q\left(\frac{1}{2}\right)$ = 0

$\therefore \quad p\left(\frac{1}{2}\right)^2+5\left(\frac{1}{2}\right)+r=0$

∴ $\frac{1}{4}$p + $\frac{5}{2}$ + r = 0 ........(ii)

$\therefore \quad 4 p+10+r=\frac{1}{4} p+\frac{5}{2}+r \quad \ldots[$ From (i) and (ii) $]$

$\begin{array}{ll}\therefore & 4 p-\frac{1}{4} p=\frac{5}{2}-10 \\ \therefore & \frac{15}{4} p=\frac{-15}{2}\end{array}$

∴ p = -2 ....(iii)

Substituting p = -2 in (i), we get

4(- 2) + 10 + r = 0

∴ -8 + 10 + r = 0

∴ r = - 2 ....(iv)

∴ p = r ... [From (iii) and (iv)]