Question

Guys please answer it if you can.​

Answer 1

Answer:

which class are you study?

Answer 2

Given -

$\alpha \: and \: \beta \: are \: the \: roots \: of \: the \: equations \: a {x}^{2} - bx + c = 0$

To prove -

$\sqrt{ \frac{\alpha}{\beta} } + \sqrt{ \frac{\beta}{\alpha} } - \sqrt{ \frac{b}{a} } = 0$

Proof-

$a {x}^{2} - bx + c = 0$

$\alpha \: \: \: and \: \beta are \: the \: roots$

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

$\alpha + \beta = \frac{b}{a} .....1)$

And,

$\alpha \times \beta = \frac{b}{a} .....2)$

Now,

L. H. S

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

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

Now taking L. C. M

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

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

$= \frac{ \frac{b}{a} }{ \sqrt{ \frac{b}{a} } } - \sqrt{ \frac{b}{a} } (by \: 1 \: and \: 2)$

$= \frac{ \sqrt{ \frac{b}{a} } \sqrt{ \frac{b}{a} } }{ \sqrt{ \frac{b}{a} } } - \sqrt{ \frac{b}{a} }$

$= \sqrt{ \frac{b}{a} } - \sqrt{ \frac{b}{a} }$

$= 0$

R. H. S

Hence Proved