Question

If $a,b,c,d$ are roots of polynomial $x^4+5x^3-4x^2+6=0$ then the value of $\frac{1}{a^2-5a+6}+\frac{1}{b^2-5b+6}+\frac{1}{c^2-5c+6}+\frac{1}{d^2-5d+6}$ is

• A. $\frac{-76}{46}$
• B. $\frac{-219}{186}$
• C. $0$
• D. $\frac{677}{1426}$

Answer 1

The correct option is D $\frac{677}{1426}$

Given the equation $f\left(x\right)=x^4+5x^3-4x^2+6=0$ and roots $a,b,c,d$

We are asked to find out the value of $\frac{1}{a^2-5a+6}+\frac{1}{b^2-5b+6}+\frac{1}{c^2-5c+6}+\frac{1}{d^2-5d+6}$

Using partial fractions, we simplify that into $\sum _{k=a}^d\frac{1}{k-3}-\sum _{k=a}^d\frac{1}{k-2}$

If the a,b,c,d are the roots of f(x), then the equation having roots as $a-3,b-3,c-3,d-3$ is $f(x+3)$

If $a-3,b-3,c-3,d-3$ are the roots of $f(x+3)$, then the equation having roots as $\frac{1}{a-3},\frac{1}{b-3},\frac{1}{c-3},\frac{1}{d-3}$ will be $f\left(\frac{1}{x+3}\right)$

By substituting $\frac{1}{x+3}$ instead of $x+3$ in $f\left(x\right)$

$f\left(\frac{1}{x+3}\right)=186x^4+219x^3+95x^2+17x+1=0$ ...... $\left(1\right)$

$\sum _{k=a}^d\frac{1}{k-3}=-\frac{219}{186}.........\left(2\right)$

In the same way, to find the $\sum _{k=a}^d\frac{1}{k-2}$ we will substitute $\frac{1}{x+2}$ in place of $x$

$f\left(\frac{1}{x+2}\right)=46x^4+76x^3+50x^2+13x+1=0$

$\sum _{k=a}^d\frac{1}{k-2}=-\frac{76}{46}=-\frac{38}{23}..........\left(3\right)$

Subtracting equation $\left(3\right)$ from equation $\left(2\right)$

$\sum _{k=a}^d\frac{1}{k-3}-\sum _{k=a}^d\frac{1}{k-2}=-\frac{219}{186}+\frac{68}{23}=\frac{2031}{4278}=\frac{677}{1426}$