Question

If r and s are zeroes of the polynomial $t^2-4t+3$, then $\frac{1}{r}+\frac{1}{s}-2rs+\frac{14}{3}$ is equal to

• A. 0
  
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (A)

0

Quadratic Equation
If $\alpha$ and $\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$
then $\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$
$f\left(t\right)=t^2-4t+3$

$r+s=4$

$rs=3$

Now,

$\frac{1}{r}+\frac{1}{s}-2rs+\frac{14}{3}=\frac{r+s}{rs}-2rs+\frac{14}{3}$

$=\frac{4}{3}-6+\frac{14}{3}=\frac{4-18+14}{3}=0$