Question

if x=root3+root2/root3-root2 and y=root 3-root 2/root3+root 2, find x square + y square

Answer 1

X =(√3+√2)/(√3-√2)

By taking conjugate, we get

X  = (√3+√2) (√3+√2) /(√3-√2) (√3+√2)

    =(√3+√2)²/(√3²-√2²)

    =3+2√6+2 / (3-2) = 5+2√6

Y   = (√3-√2)/(√3+√2)

By taking conjugate we get

Y    = (√3-√2)²/ (√3+√2)(√3-√2)

      = 3-2√6+2/√3²-√2²

       = 5-2√6/(3-2) = 5-2√6

X²     = (5+2√6)² = 25+20√6+24 = 49+20√6

Y²      = (5-2√6)² = 25-20√6+24  = 49-20√6

X²+Y² =  49+20√6+49-20√6

X²+Y² = 98

Answer 2

Answer:

Concept:

The elimination of radicals from an algebraic fraction's denominator is known as root rationalization in elementary algebra. Then, multiply by and expand the product in the denominator as in the previous step.

Step-by-step explanation:

Given:

     $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

      $y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Find:

        $x^{2}+y^{2}$

Solution:

By rationalizing x and y

$\begin{gathered}\text { If } x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \text { and } y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\\\x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=\frac{(3+2+2 \sqrt{6})}{1}=5+2 \sqrt{6} \\\end{gathered}$

$y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}=\frac{3+2-2 \sqrt{6}}{1}=5-2 \sqrt{6}$

$\\x^{2}+y^{2}=(5+2 \sqrt{6})^{2}+(5-2 \sqrt{6})^{2}\\\\={25+24+2 \cdot 5 \cdot2 \sqrt{6}+25+24-2 \cdot 5 \cdot \sqrt{6}}={49+49}=98$

                                    $x^{2}+y^{2} = 98$

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