Question

if alpha and beta are the zeros of the polynomial of a^2-5x+6, find the 1/alpha^2 +1/beta^2​

Answer 1

Let,

$f(x) = {x}^{2} - 5x + 6$

Therefore,

${x}^{2} - 5x + 6 = 0 \\ = > {x}^{2} - 2x - 3x + 6 = 0 \\ = > x(x - 2) - 3(x - 2) = 0 \\ = > (x - 3)(x - 2) = 0 \\ = > x = 3 \: \: or \: \: x = 2$

Hence,

$\alpha = 3 \: \: and \: \: \beta = 2$

So,

$\frac{1}{ { \alpha }^{2} } + \frac{1}{ { \beta }^{2} } \\ = \frac{1}{9} + \frac{1}{4} \\ = \frac{13}{36}$

HOPE THIS COULD HELP!!!

Answer 2

let

$f(x) = {x}^{2} - 5x - + 6$

so,

sum of zeros=

$\alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 5)}{1} = 5$

Product of zeros=

$\alpha \times \beta = \frac{c}{a} = \frac{6}{1} = 6$

Now,

$\frac{1}{ { \alpha }^{2} } + \frac{1}{ { \beta }^{2} } = \frac{ { \alpha }^{2} + { \beta }^{2} }{ { \alpha }^{2} \times { \beta }^{2} } = \frac{ { (\alpha + \beta )}^{2} - 2 \alpha \beta }{ { \alpha }^{2} \times { \beta }^{2} } = \frac{ { (\alpha + \beta ) }^{2} }{ { \alpha }^{2} \times { \beta }^{2} } - \frac{2}{ \alpha \beta } = \frac{ {5}^{2} }{ {6}^{2} } - \frac{2}{6} = \frac{25}{36} - \frac{2}{6} = \frac{25 - 12}{36} = \frac{13}{36}$