Question

If x=(√3+√2) / (√3-√2) and y = (√3-√2) / (√3+√2), then find the value of x2 + y2
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Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Value \ of \ x^{2}+y^{2} \ is \frac{386}{25}}$

$\bold\orange{Given:}$

$\bold{x=\frac{(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)}}$

$\bold{y=\frac{(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)}}$

$\bold\pink{To \ find:}$

Value of $\bold{x^{2}+y^{2}}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{x=\frac{(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)}}$

Rationalising the denominator

$\bold{x=\frac{(\sqrt3+\sqrt2)×(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)×(\sqrt3+\sqrt2)}}$

$\bold{x=\frac{(\sqrt3+\sqrt2)^{2}}{(3^{2}-2^{2})}}$

$\bold{x=\frac{9+2\sqrt6+4}{9-4}}$

$\bold{x=\frac{13+2\sqrt6}{5}}$

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$\bold{y=\frac{(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)}}$

Rationalising the denominator

$\bold{y=\frac{(\sqrt3-\sqrt2)×(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)×(\sqrt3-\sqrt2)}}$

$\bold{y=\frac{(\sqrt3-\sqrt2)^{2}}{(3^{2}-2^{2})}}$

$\bold{y=\frac{9-2\sqrt6+4}{9-4}}$

$\bold{y=\frac{13-2\sqrt6}{5}}$

_______________________________

Now,

x+y=$\bold{\frac{13+2\sqrt6}{5}+\frac{13-2\sqrt6}{5}}$

x+y=$\bold{\frac{26}{5}}$

Squaring both the sides

$\bold{(x+y)^{2}=\frac{676}{25}...(1)}$

_______________________________

Now,

xy=$\bold{\frac{13+2\sqrt6}{5}×\frac{13-2\sqrt6}{5}}$

xy=$\bold{\frac{13^{2}-2\sqrt6^{2}}{25}}$

xy=$\bold{\frac{169-24}{25}}$

xy=$\bold{\frac{145}{25}...(2)}$

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By identity

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

From (1) and (2)

$\bold{x^{2}+y^{2}=\frac{676}{25}-\frac{2×145}{25}}$

$\bold{x^{2}+y^{2}=\frac{676-290}{25}}$

$\bold{x^{2}+y^{2}=\frac{386}{25}}$

Therefore,

$\bold\purple{Value \ of \ x^{2}+y^{2} \ is \frac{386}{25}}$