Question

If x = √3 + √2 /√3-√2 and y = √3 - √2 /√3 + √2, then, find the value of x² + y².

Please answer fast!!

Answer 1

$\underline{\textbf{Given:}}$

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

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

$\underline{\textbf{To find:}}$

$\textsf{The value of}\;\mathsf{x^2+y^2}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{x+y}$

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

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

$\textsf{Using the identities,}$

$\boxed{\begin{minipage}{4cm} \\1.\;(a+b)^2=a^2+b^2+2ab\\\\2.\;(a-b)^2=a^2+b^2-2ab\\\\3.\;(a-b)(a+b)=a^2-b^2\\ \end{minipage}}$

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

$\mathsf{=\dfrac{10}{1}}$

$\implies\mathsf{x+y=10}$

$\mathsf{Now,}$

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

$\mathsf{x^2+y^2=(10)^2-2{\times}\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}{\times}\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

$\mathsf{x^2+y^2=100-2}$

$\implies\boxed{\mathsf{x^2+y^2=98}}$

Answer 2

Answer:

98 is the answer, I hope my answer is right