Question

if x = root 3 -root 2upon root 3+root 2 and y = root 3 +root 2 upon root 3 -root 2, find the value of x square + ysquare + xy

Answer 1

The value of $x^{2}+y^{2}+x y \text { is } \bold{99}$.

Given:

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

Solution:

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

Rationalizing the denominator for x (Numerator also can be rationalized but denominator is much easier for calculation purpose)

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

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

$=\frac{3+2-2 \sqrt{6}}{3-2}$

$=5-2 \sqrt{6}$

Squaring on both sides,

$x^{2}=25+24-20 \sqrt{6}\bold{=49-20 \sqrt{6}}$

Rationalizing the denominator for y

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

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

$=\frac{3+2+2 \sqrt{6}}{3-2}$

$=5+2 \sqrt{6}$

Squaring on both sides,

$y^{2}=25+24+20 \sqrt{6}\bold{=49+20 \sqrt{6}}$

Consider, $x^{2}+y^{2}+x y=49+20 \sqrt{6}+49-20 \sqrt{6}+(5+2 \sqrt{6})(5-2 \sqrt{6})$

$\bold{=98+25-24=99}$

Answer 2

Answer:

$The \:value \: of \: x^{2}+y^{2}+xy = 99$

Step-by-step explanation:

$Given ,\\</p><p>x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\,y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

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

/* We know the algebraic identities:

1) (a-b)²+(a+b)²=2(a²+b²)

2) (a-b)(a+b)=a²-b² */

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

$=\frac{2[(\sqrt{3})^{2}+(\sqrt{2})^{2}]}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

$=\frac{2(3+2)}{3-2}$

$=\frac{2\times 5}{1}</p><p>[tex]=\frac{10}{1}$

$=10--(1)$

$ii)xy =\big(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\big)\big(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\big)\\=1--(2)$

$Now,\\</p><p>x^{2}+y^{2}+xy\\=(x+y)^{2}-xy\\=(10)^{2}-1$

\* From (1) and (2) *\

$=100-1\\=99$

Therefore,

$The \:value \: of \: x^{2}+y^{2}+xy = 99$

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