1/2-root3 +2/root5 -root3 +1/2-root5
Answers
Answer:
If x=(2+\sqrt{5}){}^{\frac{1}{2}}+(2-\sqrt{5}){}^{\frac{1}{2}}(2+
5
)
2
1
+(2−
5
)
2
1
and y=(2+\sqrt{5}){}^{\frac{1}{2}}-(2-\sqrt{5}){}^{\frac{1}{2}}(2+
5
)
2
1
−(2−
5
)
2
1
Then evaluate x{}^{2}+y{}^{2}x
2
+y
2
Formula used :
1)(x + y) {}^{2} = x {}^{2} + y {}^{2} - 2xy1)(x+y)
2
=x
2
+y
2
−2xy
2)(x - y) {}^{2} = x {}^{2} + y {}^{2} - 2xy2)(x−y)
2
=x
2
+y
2
−2xy
3)x {}^{2} - y {}^{2} = (x + y)(x - y)3)x
2
−y
2
=(x+y)(x−y)
Solution :
we to find the value of x{}^{2}+y{}^{2}x
2
+y
2
x=(2+\sqrt{5}){}^{\frac{1}{2}}+(2-\sqrt{5}){}^{\frac{1}{2}}(2+
5
)
2
1
+(2−
5
)
2
1
Now squaring on both sides
x{}^{2} = (( \sqrt{2 + \sqrt{5} }) +( \sqrt{2 - \sqrt{5} } )) {}^{2}x
2
=((
2+
5
)+(
2−
5
))
2
x { }^{2} = 2 + \sqrt{5} + 2 - \sqrt{5} + 2 \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} }x
2
=2+
5
+2−
5
+2
2+
5
×
2−
5
x{}^{2}= 4 + 2 \times \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} } ...(1)x
2
=4+2×
2+
5
×
2−
5
...(1)
y=(2+\sqrt{5}){}^{\frac{1}{2}}-(2-\sqrt{5}){}^{\frac{1}{2}}(2+
5
)
2
1
−(2−
5
)
2
1
Squaring on both sides
y {}^{2} = (( \sqrt{2 + \sqrt{5} } ) - \sqrt{2 - \sqrt{5} } )) {}^{2}y
2
=((
2+
5
)−
2−
5
))
2
y {}^{2} = 2 + \sqrt{5} + 2 - \sqrt{5} - 2 \times \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} }y
2
=2+
5
+2−
5
−2×
2+
5
×
2−
5
y{}^{2} = 4 - 2 \times \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} } ...(2)y
2
=4−2×
2+
5
×
2−
5
...(2)
___________________________
Now add equation (1)&(2)
x {}^{2} + y {}^{2} = 4 - 2 \times \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} } + 4 - 2 \times \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} }x
2
+y
2
=4−2×
2+
5
×
2−
5
+4−2×
2+
5
×
2−
5
\implies \: x {}^{2} + y {}^{2} = 8⟹x
2
+y
2
=8