√x+3-4√(x-1)+√x+8-6√(x-1)=1. solve for x
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Answer:
Let
u
=
x
−
1
. We can then rewrite the left hand side of the equation as
√
u
+
4
−
4
√
u
+
√
u
+
9
−
6
√
u
=
√
(
√
u
−
2
)
2
+
√
(
√
u
−
3
)
2
=
∣
∣
√
u
−
2
∣
∣
+
∣
∣
√
u
−
3
∣
∣
Note the presence of
√
u
in the equation and that we are only looking for real values, so we have the restriction
u
≥
0
. With that, we will now consider all remaining cases:
Case 1:
0
≤
u
≤
4
∣
∣
√
u
−
2
∣
∣
+
∣
∣
√
u
−
3
∣
∣
=
1
⇒
2
−
√
u
+
3
−
√
2
=
1
⇒
−
2
√
u
=
−
4
⇒
√
u
=
2
⇒
u
=
4
Thus
u
=
4
is the only solution in the interval
[
0
,
4
]
Case 2:
4
≤
u
≤
9
∣
∣
√
u
−
2
∣
∣
+
∣
∣
√
u
−
3
∣
∣
=
1
⇒
√
u
−
2
+
3
−
√
u
=
1
⇒
1
=
1
As this is a tautology, every value in
[
4
,
9
]
is a solution.
Case 3:
u
≥
9
∣
∣
√
u
−
2
∣
∣
+
∣
∣
√
u
−
3
∣
∣
=
1
⇒
√
u
−
2
+
√
u
−
3
=
1
⇒
2
√
u
=
6
⇒
√
u
=
3
⇒
u
=
9
Thus
u
=
9
is the only solution in the interval
[
9
,
∞
)
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