Answers

  • Brainly User
2014-03-01T13:24:04+05:30
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)

1 5 1
2014-03-01T13:52:23+05:30
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 4q
²=25rp
hence (c) is the correct option

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