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Secondary School Math 5 points

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

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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

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