Question

if alpha bita are the zeros of the polynomial 2x^2 -4x +5 find 1/alpha ^2 + 1/bita^2​

Answer 1

Answer:

Step-by-step explanation:

Given,

$\alpha , \beta :-$

are the zeroes of the polynomial :-

$2x^2 - 4x + 5$

To Find :-

$\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2}$

Solution :-

We know that , Sum of the roots = -b/a

Product of the roots = c/a

Comparing 2x^2 - 4x + 5 with general form of quadratic equation "ax^2 + bx + c"

$\impies a = 2,b=-4,c=5$

$roots = \alpha,\beta$

$\implies \alpha + \beta = \dfrac{-(-4)}{2}$

$\alpha+\beta = \dfrac{4}{2}\\\\= 2$

[Since, by the formula of sum of the roots]

$\alpha\times \beta = \dfrac{5}{2}$

[since, by the formula of product of the roots]

As we need to find :- $\sf\dfrac{1}{\alpha^2}+\dfrac{1}{\beta}^2$

$= \dfrac{\beta^2 + \alpha^2}{\alpha^2\times\beta^2}$

$= \dfrac{\beta^2 + \alpha^2}{(\alpha\beta)^2}$

We know that , (a + b)^2 = a^2 + 2ab + b^2

a^2  + b^2 = (a + b)^2 - 2ab

$\implies \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$

$\dfrac{ (\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$

$\dfrac{(2)^2 - 2\bigg(\dfrac{5}{2}\bigg)}{\bigg(\dfrac{5}{2}\bigg)^2}$

$\dfrac{4 - 5}{\dfrac{25}{4}}$

$\dfrac{-1}{\dfrac{25}{4}}$

$\dfrac{-1\times 4}{25}$

$\dfrac{-4}{25}$

Since ,

$\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{-4}{25}$