If X= √3 - √2 / √3 + √2
and y = √3 +√2 / √3 - √2
then find x² + y² + xy
Answers
Answer:
x
2
+xy+y
2
=99
Step-by-step explanation:
Given : x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}x=
3
−
2
3
+
2
and y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}y=
3
+
2
3
−
2
We have to find x^2+xy+y^2x
2
+xy+y
2
First we calculate x^2x
2
Consider x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}x=
3
−
2
3
+
2
We first rationalize the denominator by multiply and divide by {\sqrt{3}+\sqrt{2}}
3
+
2
we get,
x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}x=
3
−
2
3
+
2
×
3
+
2
3
+
2
Simplify, we get,
\begin{gathered}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\end{gathered}
x=
(
3
)
2
−(
2
)
2
(
3
+
2
)
2
x=
3−2
(
3
+
2
)
2
x=(
3
+
2
)
2
Thus, squaring both side we get,
x^2=((\sqrt{3}+\sqrt{2})^2)^2x
2
=((
3
+
2
)
2
)
2
using algebraic identity (a+b)^2=a^2+b^2+2ab(a+b)
2
=a
2
+b
2
+2ab , we have,
x^2=(3+2+2\sqrt{6})^2=(5+2\sqrt{6})^2x
2
=(3+2+2
6
)
2
=(5+2
6
)
2
Similarly, for y^2y
2
Consider y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}y=
3
+
2
3
−
2
We first rationalize the denominator by multiply and divide by {\sqrt{3}-\sqrt{2}}
3
−
2
we get,
x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}x=
3
+
2
3
−
2
×
3
−
2
3
−
2
Simplify, we get,
y=(\sqrt{3}-\sqrt{2})^2y=(
3
−
2
)
2
y^2=((\sqrt{3}-\sqrt{2})^2)^2y
2
=((
3
−
2
)
2
)
2
using algebraic identity (a-b)^2=a^2+b^2-2ab(a−b)
2
=a
2
+b
2
−2ab , we have,
y^2=(3+2-2\sqrt{6})^2=(5-2\sqrt{6})^2y
2
=(3+2−2
6
)
2
=(5−2
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})^2x
2
+xy+y
2
=(5+2
6
)
2
+(5+2
6
)(5−2
6
)+(5−2
6
)
2
Simplify , we get,
\begin{gathered}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\end{gathered}
x
2
+xy+y
2
=25+24+20
6
+25−24+25+24−20
6
x
2
+xy+y
2
=25+24+25+25
x
2
+xy+y
2
=99
Thus, x^2+xy+y^2=99x
2
+xy+y
2
=99
Answer:
20 is the answer
the explanation is given above
hope it helps
