Question

If $\alpha ,\beta$ are the zeroes of the quadratic polynomial
$f\left(x\right)=x^2-5x+4$, then $\frac{1}{\alpha }+\frac{1}{\beta }-2\alpha \beta =$

• A. $\frac{27}{4}$
• B. $-\frac{27}{4}$
• C. $\frac{4}{27}$
• D. $-\frac{4}{27}$

Answer 1

Answer: (B)

$-\frac{27}{4}$

We know that for any

$p\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$ is a quadratic polynomial

Let $\alpha ,\beta$ be the zeroes of the polynomial $p\left(x\right)$

We know that,

$\alpha +\beta =\frac{-b}{a}$ and $\alpha \times \beta =\frac{c}{a}$

Given that

$p\left(x\right)=x^2-5x+4$

$\alpha ,\beta$ be the roots of this polynomial.

$\therefore$ Sum of the roots $=\alpha +\beta =5$

Product of the roots $=\alpha \times \beta =4$

Now, $\frac{1}{\alpha }+\frac{1}{\beta }$ $-2\alpha \beta$

$=\frac{\alpha +\beta }{\alpha \beta }$ $-2\alpha \beta$

$=\frac{5}{4}$ $-8$

$=\frac{-27}{4}$