Question

if a is equal to root 3 + root 2 by root 3 minus root 2 and b is equal to root 3 minus root 2 by root 3 + root 2 find a square + b square​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Solution :-

As given :-

$a = \dfrac{ \sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$b = \dfrac{ \sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

Now by rationalising denominator of a and b

$a = \dfrac{ \sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \dfrac{ \sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

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

$a = \dfrac{ 3 + 2\sqrt{6} + 2}{3 - 2}$

$a = 5 + 2\sqrt{6}$

$b = \dfrac{ \sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times \dfrac{ \sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$b = \dfrac{ (\sqrt{3} - \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$

$b = \dfrac{ 3 - 2\sqrt{6} + 2}{3 - 2}$

$b = 5 - 2\sqrt{6}$

Now as we know that

x² + y² = (x + y)² - 2xy

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

$= ( 5 + 2\sqrt{6} + 5 - 2\sqrt{6})^2 - 2( 5 + 2\sqrt{6})( 5 - 2\sqrt{6})$

$= (10)^2 - 2 \left( 5^2 - (2\sqrt{6})^2\right)$

$= 100 - 2(25 - 24)$

$= 100 - 2$

$= 98$

So

$\huge{\boxed{\sf{ a^2 + b^2 = 98}}}$