Question

If $-1,-1,\alpha ,\beta$ are roots of $x^4+3x^2+bx+c=0,$ then $(1+\alpha )(1+\beta )$ is equal to

• A. $0$
• B. $9$
• C. $18$
• D. $-27$

Answer 1

The correct option is B $9$

$x^4+3x^2+bx+c=(x+1)(x+1)(x-\alpha )(x-\beta )=(x+1{)}^2(x-\alpha \left)\right(x-\beta )$
Differentiating both sides
$4x^3+6x+b=(x+1{)}^2(x-\beta )+(x+1{)}^2(x-\alpha )+2(x+1)(x-\alpha )(x-\beta )$
Again differentiating, we get
$12x^2+6=(x+1{)}^2+2(x+1\left)\right(x-\beta )+(x+1{)}^2+2(x+1)(x-\alpha )+2(x+1)(x-\alpha )+2(x+1)(x-\beta )+2(x-\alpha )(x-\beta )$
Putting $x=-1,$ we get
$\Rightarrow 18=0+0+0+0+0+0+2(-1-\alpha )(-1-\beta )$
$\Rightarrow 18=2(1+\alpha )(1+\beta )$
$\therefore (1+\alpha )(1+\beta )=9$