Question

If one root of the equation $4x^3-x^2-16x+4=0$ is $8$ times the other, then the roots are

• A. $1,2,8$
• B. $\frac{1}{2},4,-1$
• C. $\frac{1}{4},-2,2$
• D. $2,4,16$

Answer 1

Answer: (C)

$\frac{1}{4},-2,2$

Let the roots of given equation $4x^3-x^2-16x+4=0$ is $p,q,r$

Let $p=8q$

Roots are $8q,q,r$

Relation betwen roots and coefficients are,

Sum of the roots$=\frac{-b}{a}$

$\therefore 8q+q+r=9q+r=\frac{-b}{a}$ $=\frac{-(-1)}{4}=\frac{1}{4}$

Product of the roots$=\frac{-d}{a}$

$8q\times q\times r=8q^{2}r=\frac{-d}{a}=\frac{-4}{4}=-1$

Sum of products of roots$=\frac{c}{a}$

$8q\times q+8q\times r+q\times r=\frac{c}{a}=\frac{-16}{4}=-4$

Solving $8q^2(\frac{1}{4}-9q)=-1$

we get $q=1/4$

and $r=-2$

So roots are $8\times \frac{1}{4},\frac{1}{4},-2=2,\frac{1}{4},-2$