Question

If $-3,6,9$ are the roots of $p{x}^3+q{x}^2+rx+s=0$, then the roots of $27p{x}^3+9q{x}^2+3rx+s=0$ are

• A. $-1,2,3$
• B. $-3,6,9$
• C. $-2,4,6$
• D. $-\frac{1}{3},\frac{2}{3},1$

Answer 1

The correct option is A $-1,2,3$

As $-3,6,9$ are the roots of $p{x}^3+q{x}^2+rx+s=0,$ then $S_1=-3+6+9=12=-\frac{q}{p}\Rightarrow \frac{q}{p}=-12$

$S_2=-3\times 6+(-3)\times 9+6\times 9=\frac{r}{p}$

$\Rightarrow -18-27+54=\frac{r}{p}\Rightarrow \frac{r}{p}=9$

$S_3=-3\times 6\times 9=-\frac{s}{p}\Rightarrow 162=\frac{s}{p}$

$\therefore 27p{x}^3+9q{x}^2+3rx+s=0$

$\Rightarrow x^3+\frac{q}{3p}{x}^2+\frac{rx}{9p}+\frac{s}{27p}=0$

$\Rightarrow x^3+\frac{(-12)}{3}{x}^2+\frac{9x}{9}+\frac{162}{27}=0$

$\Rightarrow x^3-4x^2+x+6=0$

$\Rightarrow (x+1)(x-2)(x-3)=0$

Hence, the roots are $-1,2,3$.