Question

if x=√3+√2/√3-√2 and y=√3-√2/√3+√2,find the value of x²+xy+y²

Answer 1

Answer:

$x^2+xy+y^2=99$

Step-by-step explanation:

Given :  $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

We have to find $x^2+xy+y^2$

First we calculate $x^2$

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

We first rationalize the denominator  by multiply and divide by ${\sqrt{3}+\sqrt{2}}$

we get,

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

Simplify, we get,

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

Thus, squaring both side we get,

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

using algebraic identity $(a+b)^2=a^2+b^2+2ab$ , we have,

$x^2=(3+2+2\sqrt{6})^2=(5+2\sqrt{6})^2$

Similarly, for $y^2$

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

We first rationalize the denominator  by multiply and divide by ${\sqrt{3}-\sqrt{2}}$

we get,

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

Simplify, we get,

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

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

using algebraic identity $(a-b)^2=a^2+b^2-2ab$ , we have,

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

then $x^2+xy+y^2=(5+2\sqrt{6})^2+(5+2\sqrt{6})(5-2\sqrt{6})+(5-2\sqrt{6})^2$

Simplify , we get,

$x^2+xy+y^2=25+24+20\sqrt{6}+25-24+25+24-20\sqrt{6}\\\\ x^2+xy+y^2=25+24+25+25\\\\ x^2+xy+y^2=99$

Thus, $x^2+xy+y^2=99$

Answer 2

Answer:

First rationalise X and Y

X = (✓3-✓2)² / 3-2 = (✓3-✓2)² / 1 = (✓3-✓2)²

Y = (✓3+✓2)² / 3-2 = (✓3+✓2)² / 1 = (✓3+✓2)²

X²+Y²+XY = ( (✓3-✓2)²)² + ( (✓3+✓2)²)² + (✓3-✓2)²* (✓3+✓2)²