Not so easy bros
Find x if X > 0, and x is an integer
( 1 2 x + 6 ) ( 6 x + 3 ) ( 2 x + 1 ) ( 4 x + 2 ) = 2 2 5 0 0
The answer is 2.
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2 solutions
if you calculate properly,
(
1
2
x
+
6
)
(
2
x
+
1
)
=
(
6
x
+
3
)
(
4
x
+
2
)
so, we can write :
(
1
2
x
+
6
)
(
2
x
+
1
)
(
6
x
+
3
)
(
4
x
+
2
)
=
(
1
2
x
+
6
)
(
2
x
+
1
)
2
=
2
2
5
0
0
(
1
2
x
+
6
)
(
2
x
+
1
)
=
2
2
5
0
0
(
1
2
x
+
6
)
(
2
x
+
1
)
=
1
5
0
2
4
x
2
+
2
4
x
+
6
=
1
5
0
2
4
x
2
+
2
4
x
=
1
4
4
x
2
+
x
=
6
since x is an integer and
X
2
>
X
3+3 =6
_ 4+2=6__
5+1=6
ONLY THE SECOND ONE WORKS SO THEREFORE X= 2
Much simpler way,
6 ( 2 x + 1 ) 3 ( 2 x + 1 ) ( 2 x + 1 ) 2 ( 2 x + 1 ) = 2 2 5 0 0
( 2 x + 1 ) 4 = 6 2 5 = 5 4
x = 2
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Yup. Did the same way! Makes this problem a one-liner. :D
Actually, you can factor the LHS as 36(2x+1)^4, it makes the equation: 36(2x+1)^4=22500. Now (1) Divide both sides by 36 to get (2x+1)^4 = 625. (2) Take 4th root to both sides to get 2x+1=±5 and 2x+1=±5i. (3) Solve the equation for x using each root (-5, 5, - 5i and 5i), the only positive integer solution is x = 2.