Question

If $a,b,c\in R$ and the equations $a{x}^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two roots in common, then

• A. $a=b\ne c$
• B. $a=b=-c$
• C. $a=b=c$
• D. None of these

Answer 1

The correct option is C $a=b=c$

We have, $x^3+3x^2+3x+2=0$

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

$\Rightarrow x=-2,\frac{-1\pm \sqrt{3}i}{2}\Rightarrow x=-2,\omega ,{\omega }^2$

Since $a,b,c\in R,a{x}^2+bx+c=0$ cannot have one real and one imaginary root.

Therefore, two common roots of $a{x}^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ are $\omega ,{\omega }^2$.

Thus, $-\frac{b}{a}=\omega +{\omega }^2=-1$

$\Rightarrow a=b$ and $\frac{c}{a}=\omega {\omega }^2=1\Rightarrow c=a$

$\Rightarrow a=b=c$