Solve
Guides
Join / Login
Use app
Login
0
You visited us
0
times! Enjoying our articles?
Unlock Full Access!
Standard VIII
Mathematics
Question
x
=
2
√
3
−
1
2
√
3
+
1
,
y
=
2
√
3
+
1
2
√
3
−
1
.
Find
x
2
−
y
2
Open in App
Solution
Verified by Toppr
Considering the question to be,
x
⇒
2
√
3
−
1
2
√
3
+
1
y
⇒
2
√
3
+
1
2
√
3
−
1
x
2
=
13
−
4
√
3
13
+
4
√
3
y
2
=
13
+
4
√
3
13
−
4
√
3
x
2
−
y
2
=
(
13
−
4
√
3
13
+
4
√
3
)
2
−
(
13
+
4
√
3
13
−
4
√
3
)
2
=
−
2
×
2
×
13
×
4
√
3
121
=
208
√
3
121
x
2
−
y
2
=
208
√
3
121
Was this answer helpful?
0
Similar Questions
Q1
x
=
2
√
3
−
1
2
√
3
+
1
,
y
=
2
√
3
+
1
2
√
3
−
1
.
Find
x
2
−
y
2
View Solution
Q2
Show that:
x
2
3
√
y
−
2
y
2
√
x
−
2
×
y
2
√
x
−
2
√
x
4
3
=
1
y
View Solution
Q3
The value of
1
x
2
+
1
y
2
, where
x
=
2
+
√
3
and
y
=
2
−
√
3
, is?
View Solution
Q4
If
x
=
√
3
+
√
2
,
y
=
1
√
3
+
√
2
then find
(
x
+
y
)
2
−
(
x
−
y
)
2
.
View Solution
Q5
Let a solution
y
=
y
(
x
)
of the differential equation
x
√
x
2
−
1
d
y
−
y
√
y
2
−
1
d
x
=
0
satisfy
y
(
2
)
=
2
√
3
STATEMENT-1:
y
(
x
)
=
sec
(
s
e
c
−
1
x
−
π
6
)
STATEMENT-2 :
y
(
x
)
is given by
1
y
=
2
√
3
x
−
√
1
−
1
x
2
View Solution