Question

find value of Alpha and beta.. explanation plz​

Answer 1

Given :

$\alpha + \beta = 12$

$\alpha - \beta = 2 \sqrt{3}$

To Find :

Value of $\alpha \beta$

$\alpha + \beta = 12$

$= > \alpha = 12 - \beta ...(1)$

put alpha value into another equation:

$= > \alpha - \beta = 2 \sqrt{3}$

$= > 12 - \beta - \beta = 2 \sqrt{3}$

$= > 12 - 2 \beta = 2 \sqrt{3}$

$= > - 2 \beta = 2 \sqrt{3} - 12$

$= > \beta = \frac{2 \sqrt{3} - 12}{ - 2}$

Taking 2 common from L.H.S

$= > \beta = \frac{2( \sqrt{3} - 6) }{ - 2} = - ( \sqrt{3} - 6)$

$\bold{\boxed{\purple{\beta = 6 - \sqrt{3}}}}$

Put beta's value in equation 1

$= > \alpha = 12 - \beta$

$= > \alpha = 12 - (6 - \sqrt{3} )$

$= > \alpha = 12 - 6 + \sqrt{3}$

$\bold{\boxed{\purple{ \alpha = 6 + \sqrt{3}}}}$

$= > \alpha \beta = (6 + \sqrt{3} )(6 - \sqrt{3} )$

Here,this identity is used:-

$\bold{\boxed{{x}^{2} - {y}^{2} = (x + y)(x - y)}}$

$= > \alpha \beta = {(6)}^{2} - {( \sqrt{3}) }^{2}$

$\bold{\red{ \alpha \beta = 36 - 3 = 33}}$