Question

If α and β are solutions of a quadratic equation
ax^2+ bx + c = 0 , then find the
value of 1/α2 + 1/β2

please answer fast!!​

Answer 1

Given -

$\alpha \: and \: \beta \: are \: solutions \: \: of \: \: a \: \: quadratic \\ equation \: a {x}^{2} + bx + c = 0$

To find -

$The \: value \: \: of \: \frac{1}{ \alpha {}^{2} } + \frac{1}{ { \beta }^{2} }$

Solution -

$ax {}^{2} + bx + c = 0$

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

And,

$\alpha \beta = \frac{c}{a}$

Now,

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

By taking L. C. M we will get -

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

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

Putting the values we will get -

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

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

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

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

$= \frac{a {b}^{2} - 2 {a}^{2} }{ {a}^{3} } \times \frac{ {a}^{2} }{ {c}^{2} }$

$= \frac{a( {b}^{2} - 2a)}{a} \times \frac{1}{ {c}^{2} }$

$= {b}^{2} - 2a \times \frac{1}{ {c}^{2} }$

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

$\therefore \: The \: value \: of \: \frac{1}{ { \alpha }^{2} } + \frac{1}{ { \beta }^{2} } \: is \: \: \frac{ {b}^{2} - 2a}{c}$