Question

Plz say explaination and answer for 56th question

Answer 1

Hiii. ....friends

The answer is here,

$= > \: a {x}^{2} + bx + c = 0$

By using formula, we get.

$= > \: \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \: \: be \: \: \alpha .........(1)$

$= > \: \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \: \: \: be \: \: \: \beta ..........(2)$

And given other equation,

Let the lamba be gamma for my convience,


$= > \: a {x}^{2} + bx \gamma + c { \gamma }^{2} = 0$

Again by using formula,

$= > \: \frac{ - b \gamma + \sqrt{ {b}^{2} { \gamma }^{2} - 4ac { \gamma }^{2} } }{2a}$

$= > \: \frac{ \gamma (- b + \sqrt{ {b}^{2} - 4ac} )}{2a}...........(3)$

Similarly we get,

$= > \: \frac{ \gamma ( - b - \sqrt{ {b}^{2} - 4ac } )}{2a} .............(4)$
From equations 1 & 2 ,

The roots for the new equation is

$= > \: \gamma \alpha \: \: and \: \gamma \beta$

So, The correct option is 1.

:-)Hope it helps u.