Question

help me by finding the answer of this question.
the actual answer is 35.
but I need steps !!
find asap:)​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\bf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\,\,\,\,and\,\,\,\,y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

Now,

$\sf{x+y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

$\sf{\implies\,x+y=\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}$

$\sf{\implies\,x+y=\dfrac{2\left\{\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2\right\}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}}$

$\sf{\implies\,x+y=\dfrac{2\left\{3+2\right\}}{3-2}}$

$\sf{\implies\,x+y=10}$

And,

$\sf{xy=\left(\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right)\cdot\left(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)}$

$\sf{\implies\,xy=1}$

Now,

$\sf{\green{x^2+xy+y^2}}$

$\sf{=x^2+xy+xy+y^2-xy}$

$\sf{=x^2+2xy+y^2-xy}$

$\sf{=(x+y)^2-xy}$

$\sf{=(10)^2-1}$

$\sf{=100-1=99}$