if x*y=whole square root of x^2 +y^2 , the value of (1*2 x square root of 2 ) (1*-2 x square root of 2) is :
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Step-by-step explanation:
Explanation:
y
=
√
x
2
+
6
x
+
8
.
∴
y
2
=
x
2
+
6
x
+
8
.
Adding
1
to both sides,
y
2
+
1
=
x
2
+
6
x
+
9
=
(
x
+
3
)
2
.
Taking square root,
√
y
2
+
1
=
x
+
3
,
or
,
x
=
√
y
2
+
1
−
3
Therefore,
2
x
+
8
,
=
2
{
√
y
2
+
1
−
3
}
+
8
,
=
2
√
y
2
+
1
+
2
,
=
2
+
2
√
1
−
(
−
y
2
)
,
=
2
+
2
√
1
−
(
i
y
)
2
,
=
2
+
2
√
(
1
+
i
y
)
(
1
−
i
y
)
,
=
(
1
+
i
y
)
+
(
1
−
i
y
)
+
2
√
(
1
+
i
y
)
(
1
−
i
y
)
,
=
(
√
1
+
i
y
)
2
+
(
√
1
−
i
y
)
2
+
2
√
(
1
+
i
y
)
(
1
−
i
y
)
,
i
.
e
.
,
,
2
x
+
8
=
{
√
1
+
i
y
+
√
1
−
i
y
}
2
.
Taking square root, we get,
√
2
x
+
8
=
√
1
+
i
y
+
√
1
−
i
y
,
as desired!
Q.E.D.
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