Question

If two roots of the equation, $x^3-7x^2+4x+12=0$ are in the ratio $1:3$, then the roots are

• A. $-6,-2,-1$
• B. $-1,2,6$
• C. $1,3,4$
• D. $-2,3,6$

Answer 1

The correct option is B $-1,2,6$

Let the roots be $a,3a,b$.
The Sum of the roots is $=4a+b=7$ .. (i)
$\left(a\right)\left(3a\right)+\left(a\right)\left(b\right)+\left(3a\right)\left(b\right)=4$
$\Rightarrow 3a^2+4ab=4$ .. (ii)
The Product of roots is $3a^{2}b=-12$ .. (iii)
Substituting the value of $b$ from (i) in (ii), we get ,
$3a^2+4a(7-4a)=4$
$\Rightarrow 13a^2-28a+4=0$
$\Rightarrow (a-2)(13a-2)=0$
$\Rightarrow a=2$ or $a=\frac{2}{13}$
For $a=2$ , $b=-1$ (By substituting in (i)). This satisfies equation (iii).
For $a=\frac{2}{13}$, $b=\frac{83}{13}$, this does not satisfy equation (iii).
Hence, the roots are $(-1,2,6)$.