Question

pls solve this problem​

Answer 1

Answer:

15

Step-by-step explanation:

$a = \frac{\sqrt{3}+1 }{\sqrt{3}-1 }$

Multiply numerator and denominator with $\sqrt{3}+1$

$a = \frac{(\sqrt{3} +1)^{2} }{3-1}$

$a = \frac{4+2\sqrt{3} }{2}$

$a = 2 + \sqrt{3}$

$b = \frac{1}{a}$

$b = \frac{1}{2+\sqrt{3} }$

Multiply numerator and denominator with $2-\sqrt{3}$

$b = \frac{2-\sqrt{3} }{4-3}$

$b = 2 - \sqrt{3}$

So finally, $a = 2 + \sqrt{3}$ and $b = 2 - \sqrt{3}$

So, $a^2+ab+b^2 = (a+b)^2-ab$

=> $a^2+ab+b^2$ = $(4)^2-(1)^2$

=>  $a^2+ab+b^2$ = 15