x = √2+1 / √2-1 find the value and y = √2-1 / √2+1
Answers
Answer:
find the value of what? question hi acche ni daala
Answer:
Answer:
x^2+y^2+xy=35x
2
+y
2
+xy=35
Step-by-step explanation:
Given : x=\frac{\sqrt2+1}{\sqrt2-1}x=
2
−1
2
+1
and y=\frac{\sqrt2-1}{\sqrt2+1}y=
2
+1
2
−1
To find : The value of x^2+y^2+xyx
2
+y
2
+xy
Solution :
First we solve the value of x and y be rationalizing,
The value of x,
x=\frac{\sqrt2+1}{\sqrt2-1}x=
2
−1
2
+1
x=\frac{\sqrt2+1}{\sqrt2-1}\times\frac{\sqrt2+1}{\sqrt2+1}x=
2
−1
2
+1
×
2
+1
2
+1
x=\frac{(\sqrt2+1)^2}{(\sqrt2)^2-1^2}x=
(
2
)
2
−1
2
(
2
+1)
2
x=\frac{2+1+2\sqrt2}{2-1}x=
2−1
2+1+2
2
x=3+2\sqrt2x=3+2
2
The value of y,
y=\frac{\sqrt2-1}{\sqrt2+1}y=
2
+1
2
−1
y=\frac{\sqrt2-1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1}y=
2
+1
2
−1
×
2
−1
2
−1
y=\frac{(\sqrt2-1)^2}{(\sqrt2)^2-1^2}y=
(
2
)
2
−1
2
(
2
−1)
2
y=\frac{2+1-2\sqrt2}{2-1}y=
2−1
2+1−2
2
y=3-2\sqrt2y=3−2
2
Now, Substitute the value of x and y in the expression
x^2+y^2+xyx
2
+y
2
+xy
=(3+2\sqrt2)^2+(3-2\sqrt2)^2+(3+2\sqrt2)(3-2\sqrt2)=(3+2
2
)
2
+(3−2
2
)
2
+(3+2
2
)(3−2
2
)
=9+8+12\sqrt2+9+8-12\sqrt2+3^2-(2\sqrt2)^2=9+8+12
2
+9+8−12
2
+3
2
−(2
2
)
2
=17+17+9-8=17+17+9−8
=35=35
Therefore, x^2+y^2+xy=35x
2
+y
2
+xy=35