Question

13+
If x=
√3-√2
√3+ √2
and y
+ V2
√3-√2
then the value of x² + xy + y2 is
1​

Answer 1

Step-by-step explanation:

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,