if x is equals to root 3 + 1 / root 3 - 1, Y is equals to root 3 minus 1 by root 3+1 find the value of x2 +xy-y2
Answers
Hence, the answer is found.
Step-by-step explanation:
\huge{\underline{\bold {Solution :}}}
Solution:
\begin{gathered}\sf x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \\ \\ \sf x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \times \frac{ \sqrt{3 + 1} }{ \sqrt{3} + 1 } \\ \\ \sf x = \frac{ (\sqrt{3} + 1)^{2}}{(\sqrt{3}) {}^{2} - (1)^{2} } \\ \\ \sf x = \frac{( \sqrt{3})^{2} + 1^{2} + 2 \times \sqrt{3} \times 1 }{3 - 1} \\ \\ \sf x = \frac{3 + 1 + 2 \sqrt{3} }{2} \\ \\ \sf x = \frac{4 + 2 \sqrt{3} }{2} \\ \\ \sf x = \frac{2(2 + \sqrt{3})}{2} \\ \\ \boxed{\bf x = 2 + \sqrt{3}}\end{gathered}
x=
3
−1
3
+1
x=
3
−1
3
+1
×
3
+1
3+1
x=
(
3
)
2
−(1)
2
(
3
+1)
2
x=
3−1
(
3
)
2
+1
2
+2×
3
×1
x=
2
3+1+2
3
x=
2
4+2
3
x=
2
2(2+
3
)
x=2+
3
\sf Now, \: again \: find \: value \: of \: y Now,againfindvalueofy
\begin{gathered}\sf y = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \\ \\ \sf y = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1 } \\ \\ \sf y = \frac{ (\sqrt{3} - 1)^{2}}{(\sqrt{3}) {}^{2} - (1)^{2} } \\ \\ \sf y = \frac{( \sqrt{3})^{2} + 1^{2} - 2 \times \sqrt{3} \times 1 }{3 - 1} \\ \\ \sf y = \frac{3 + 1 - 2 \sqrt{3} }{2} \\ \\ \sf y = \frac{4 - 2 \sqrt{3} }{2} \\ \\ \sf y = \frac{2(2 - \sqrt{3})}{2} \\ \\ \boxed{\bf y = 2 - \sqrt{3}}\end{gathered}
y=
3
+1
3
−1
y=
3
+1
3
−1
×
3
−1
3
−1
y=
(
3
)
2
−(1)
2
(
3
−1)
2
y=
3−1
(
3
)
2
+1
2
−2×
3
×1
y=
2
3+1−2
3
y=
2
4−2
3
y=
2
2(2−
3
)
y=2−
3
\underline{\bf To \: find \: value \: of \: x^{2} + xy - y^{2} ;}
Tofindvalueofx
2
+xy−y
2
;
\begin{gathered}\bf x^{2} +xy - y^{2} \\ \\ \sf (2 + \sqrt{3} ){}^{2} + (2 + \sqrt{3})(2 - \sqrt{3}) - (2 - \sqrt{3} ) {}^{2} \\ \\ \sf 4 + 3 + 4 \sqrt{3} + 4 - 3 - 4 - 3 + 4 \sqrt{3} \\ \\ \sf 11 + 4 \sqrt{3} - 10 + 4 \sqrt{3} \\ \\ \boxed{ \underline{ \red{ \bf 1 + 8 \sqrt{3} }}}\end{gathered}
x
2
+xy−y
2
(2+
3
)
2
+(2+
3
)(2−
3
)−(2−
3
)
2
4+3+4
3
+4−3−4−3+4
3
11+4
3
−10+4
3
1+8
3
Hence, the answer is found.