Question

The condition that one root of the quadratic equation px²+ qx+r=0 is four times the other is
(a) 4r² =25pq
(b) 4p² =25qr
(c) 4q²= 25pr
(c) 5q² = 24pr

Answer 1

let roots a and 4a
therfore $(x-4a)(x-a)= x^{2} -5a+ 4a^{2}$
compare with $x^{2} + \frac{p}{q}x+ \frac{r}{p}=0$
therfore
$-5a= \frac{q}{p}, 4a^{2}= \frac{r}{p}$
$a= \frac{q}{-5p}, a^{2}= \frac{q^{2}}{ 25p^{2} }= \frac{r}{4p}$
therefore answer is $4 q^{2}=25rp$ which is (c)

Answer 2

given,
px²+ qx+r=0 is the equation
let x,4x be the roots
then,
sum of the roots=5x=-q/p       ........1
product of the roots=4x²=r/p        ........2
substituting x value from eq 1 in eq 2,
4*(-q/5p)²=r/p
or 4q²=25rp
hence (c) is the correct option