Question

If 2,1 / 2 are the zeros of px^2+5x+r prove that p = r

Answer 1

$p(x) = px {}^{2} + 5x + r$
If 2 is the zero of p(x). So,
p(2)=0
$p \times 2 {}^{2} + 5 \times 2 + r = 0 \\ 4p + 10 + r = 0 \\ 4p + r = - 10 \\ r = - 10 - 4p \: \: ....(i)$
If 1/2 is the zero of p(x). So,
p(1/2)=0
$p( \frac{1}{2} ) {}^{2} + 5 \times \frac{1}{2} + r = 0 \\ \frac{p}{4} + \frac{5}{2} + r = 0 \\ \frac{p}{4} + r = \frac{ - 5}{2} \: \: .....(ii)$
From equation (i),
$\frac{p}{4} - 10 - 4p = \frac{ - 5}{2} \\ \frac{p - 16p}{4} = \frac{ - 5}{2} + 10 \\ \frac{ - 15p}{4} = \frac{ - 5 + 20}{2} \\ p = \frac{15 \times 4}{2 \times - 15} \\ p = - 2$
$r = - 10 - 4p \\ = - 10 - 4 \times - 2 \\ = - 10 + 8 \\ = - 2$
So, p=r= -2
----proved----