Question

If Alfa and bita are the zeroes of the quadratic polynomial x2-2x-8 find the value of 1/alfa-1/bita

Answer 1

Answer :

$\sf\dfrac{-3}{4}$

Explanation :

Given polynomial,

$= {x}^{2} - 2x - 8$

On comparing with $ax^2 + bx + c$

We have,

• a = 1
• b = -2
• c = -8

$\sf \: and \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the \: polynomial$

Here, we need to find

$= \dfrac{1}{ \alpha } - \dfrac{1}{ \beta } \\ = \dfrac{ \beta - \alpha }{ \alpha \beta }$

Here,

$\bull \: \sf \alpha \beta (product \: of \: the \: zeroes) = \dfrac{c}{a} = { - 8}$

And also,

$\bull \: \sf \beta - \alpha(difference \: between \: the \: roots) = \dfrac{ \sqrt{D} }{a}$

$\sf \: Finding \: ‘D’ :-$

$\sf = {b}^{2} - 4ac$

$\sf = {( - 2)}^{2} - 4(1)( - 8)$

$\sf = 4 + 32$

$\sf \: = 36$

So,

$\sf \to \beta - \alpha = \frac{\sqrt{36}}{1}$

$\sf \to \beta - \alpha = 6$

Now, substituting the values,

$\sf \to \: \dfrac{ \beta - \alpha }{ \alpha \beta }$

$\sf \to \: \frac{6}{-8}$

$\sf \to \: \dfrac{-3}{4}$