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. $a\ne b\ne c$
• B. $a=b\ne -c$
• C. $a=b\ne c$
• D. $a=b=c$

Answer 1

The correct option is D

$a=b=c$

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

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

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

$\Rightarrow (x+2)(x^2+x+1)=0$

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

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 $x^3+3x^2+2=0$ are $\omega ,{\omega }^2$ .

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

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

$\Rightarrow a=b=c$