Question

If $\alpha$ and $\beta$ are the roots of the quadratic equation $(x-2)(x-3)+(x-3)(x+1)+(x+1)(x-2)=0$, then the value of $\frac{1}{(\alpha +1)(\beta +1)}+\frac{1}{(\alpha -2)(\beta -2)}+\frac{1}{(\alpha -3)(\beta -3)}$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

$(x-2)(x-3)+(x-3)(x+1)+(x+1)(x-2)=0\Rightarrow x^2-5x+6+x^2-2x-3+x^2-x-2=0\Rightarrow 3x^2-8x+1=0$
The sum and product of the roots are,
$\alpha +\beta =\frac{8}{3}\text{and}\alpha \beta =\frac{1}{3}$
Now,
$\frac{1}{(\alpha +1)(\beta +1)}+\frac{1}{(\alpha -2)(\beta -2)}+\frac{1}{(\alpha -3)(\beta -3)}=\frac{1}{\alpha \beta +(\alpha +\beta )+1}+\frac{1}{\alpha \beta -2(\alpha +\beta )+4}+\frac{1}{\alpha \beta -3(\alpha +\beta )+9}=\frac{1}{4}-1+\frac{3}{4}=0$