4 x 5 + 4 8 3 3 − 9 6 3 9 = 5 x 4 − 4 8 3 3 + 9 6 3 9
If real x satisfies the equation above, then find the value of the infinite root:
3 x 3 + 3 x 3 + 3 x 3 + ⋯
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I think it will be easier start with y = that equation , we can see that y 5 + y 4 = x 5 + x 4 , so we need to show that y is a one-to-one function such that y = x is a unique solution. We're left to show that ( x 3 − 2 4 ) is a factor of the equation x = 4 x 5 + 4 8 3 3 − 9 6 3 9 . Thoughts?
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Since we have to find the fourth root of :
x 5 + 4 8 3 3 − 9 6 3 9
Therefore this expression should be positive. This means that : x 5 + 4 8 3 3 − 9 6 3 9 > 0 x 5 > 9 6 3 9 − 4 8 3 3 x 5 > 1 3 0 This means that x is positive. Let : y = 4 x 5 + 4 8 3 3 − 9 6 3 9 = 5 x 4 − 4 8 3 3 + 9 6 3 9 This will give us the 2 equations: y 4 = x 5 + 4 8 3 3 − 9 6 3 9 . . . E Q ( 1 )
And :
y 5 = x 4 − 4 8 3 3 + 9 6 3 9 . . . E Q ( 2 )
There is an important remark here: if we exchange x and y in equation 1 we shall get equation 2. That means we can use only one of the equations replacing y by x. From equation 2:
x 5 = x 4 − 4 8 3 3 + 9 6 3 9
x 5 − x 4 + 4 8 3 3 − 9 6 3 9 = 0 . . . E Q ( 3 )
Let x be :
x = a 3 3
In that case we get:
x 4 = 3 a 4 3 3 . . . E Q ( 4 )
And
x 5 = 3 a 5 3 9 . . . E Q ( 5 )
From equations 3, 4 and 5 we get:
3 a 5 3 9 − 3 a 4 3 3 + 4 8 3 3 − 9 6 3 9 = 0 . . . E Q ( 6 )
Simplifying equation 6 by:
3 3 3
we get:
a 5 3 3 − a 4 + 1 6 − 3 2 3 3 = 0
( a 5 − 3 2 ) 3 3 − ( a 4 − 1 6 ) = 0
Let b = cubic root of 3 and factoring we get :
( a − 2 ) ( a 4 + 2 a 3 + 4 a 2 + 8 a + 1 6 ) ( b ) − ( a − 2 ) ( a 3 + 2 a 2 + 4 a + 8 ) = 0
Taking (a – 2) as a common factor we get:
( a − 2 ) ( ( a 4 + 2 a 3 + 4 a 2 + 8 a + 1 6 ) ( b ) − ( a 3 + 2 a 2 + 4 a + 8 ) ) = 0
This will give us the following values of a:
a – 2 = 0 therefore a = 2.
The other values of a come from the 2nd parentheses :
( a 4 + 2 a 3 + 4 a 2 + 8 a + 1 6 ) ( b ) − ( a 3 + 2 a 2 + 4 a + 8 ) = 0 ( a 4 + ( 2 b − 1 ) a 3 + ( 4 b − 2 ) a 2 + ( 8 b − 4 ) a + ( 1 6 b − 8 ) = 0
This is an equation of the fourth degree in a where a is positive (since x is positive as shown above) and all coefficients of the equation (1, (2b – 1), (4b – 2), (8b – 4) and (16b - 8)) are all positive, which is impossible so we can conclude that the only solution is a = 2. Therefore :
x = 2 3 3
And
x 3 = 2 4
To calculate the infinite root let’s make it equal to z. we get:
z = 3 2 4 + 3 2 4 + 3 2 4 + . . . z = 3 2 4 + z
Cubing both sides:
z 3 = z + 2 4 z 3 − z − 2 4 = 0
Factoring we get:
( z − 3 ) ( z 2 + 3 z + 8 ) = 0
The first parentheses gives z = 3 and the second parentheses give imaginary roots so it should be rejected.
A n s w e r = 3 .