Question

if alpha and beta are the zeros of polynomial f (x) ax square + bx + c then find one by Alpha square + 1 by beta square

Answer 1

Let x and y are the roots for f(x)=ax square+bx+c.

Then1/xsquare +1/ysquare = xsquare+ysquare/xywholesquare

=x+y wholesquare-2xy/xywholesquare

=in numerator-b/a-2c/a,in denominator c/a

=(-b-2c/a)(a/c)

=-b-2c/c

Answer 2

Answer:

The value of  $\frac{1}{\alpha ^{2} } + \frac{1}{\beta ^{2} }$ is $\frac{b^{2} -2ac }{c^{2} }$.

Step-by-step explanation:

Given :  α and β are the zeroes of the polynomial f(x) = ax² + bx + c.

To find: $\frac{1}{\alpha ^{2} } + \frac{1}{\beta ^{2} }$

Concept: If α and β are the zeroes of the polynomial f(x) = ax² + bx + c;

                α+β = -b/a

                αβ = c/a

Solution: α and β are the zeroes of the polynomial f(x) = ax² + bx + c;

               So, α+β = -b/c ; αβ = c/a

Now,

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

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

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

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

= $\frac{(\frac{-b}{a})^{2} -\frac{2c}{a} }{(\frac{c}{a} ){2} }$

= $\frac{b^{2} -2ac }{c^{2} }$

Thus, the value of  $\frac{1}{\alpha ^{2} } + \frac{1}{\beta ^{2} }$ is $\frac{b^{2} -2ac }{c^{2} }$.

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