Roots Everywhere

Algebra Level 5

x 5 + 48 3 3 96 9 3 4 = x 4 48 3 3 + 96 9 3 5 \sqrt[4] {x ^ 5 + 48 \sqrt[3] {3} - 96 \sqrt[3] {9}} = \sqrt[5] {x ^ 4 - 48 \sqrt[3] {3} + 96 \sqrt[3] {9}}

If real x x satisfies the equation above, then find the value of the infinite root:

x 3 + x 3 + x 3 + 3 3 3 \sqrt[3] {x ^ 3 + \sqrt[3] {x ^ 3 + \sqrt[3] {x ^ 3 + \cdots } } }


The answer is 3.

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1 solution

Karim Fawaz
Aug 18, 2016

Since we have to find the fourth root of :

x 5 + 48 3 3 96 9 3 x ^ 5 + 48 \sqrt[3] {3} - 96 \sqrt[3] {9}

Therefore this expression should be positive. This means that : x 5 + 48 3 3 96 9 3 > 0 x ^ 5 + 48 \sqrt[3] {3} - 96 \sqrt[3] {9} > 0 x 5 > 96 9 3 48 3 3 x ^ 5 > 96 \sqrt[3] {9} - 48 \sqrt[3] {3} x 5 > 130 x ^ 5 > 130 This means that x is positive. Let : y = x 5 + 48 3 3 96 9 3 4 = x 4 48 3 3 + 96 9 3 5 y = \sqrt[4] {x ^ 5 + 48 \sqrt[3] {3} - 96 \sqrt[3] {9}} = \sqrt[5] {x ^ 4 - 48 \sqrt[3] {3} + 96 \sqrt[3] {9}} This will give us the 2 equations: y 4 = x 5 + 48 3 3 96 9 3 . . . E Q ( 1 ) y ^ 4 = x ^ 5 + 48 \sqrt[3] {3} - 96 \sqrt[3] {9} ... EQ (1)

And :

y 5 = x 4 48 3 3 + 96 9 3 . . . E Q ( 2 ) y ^ 5 = x ^ 4 - 48 \sqrt[3] {3} + 96 \sqrt[3] {9} ... EQ (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 48 3 3 + 96 9 3 x ^ 5 = x ^ 4 - 48 \sqrt[3] {3} + 96 \sqrt[3] {9}

x 5 x 4 + 48 3 3 96 9 3 = 0... E Q ( 3 ) x ^ 5 - x ^ 4 + 48 \sqrt[3] {3} - 96 \sqrt[3] {9} = 0 ... EQ (3)

Let x be :

x = a 3 3 x = a\sqrt[3] {3}

In that case we get:

x 4 = 3 a 4 3 3 . . . E Q ( 4 ) x ^ 4 = 3a ^ 4 \sqrt[3] {3}... EQ (4)

And

x 5 = 3 a 5 9 3 . . . E Q ( 5 ) x ^ 5 = 3a ^ 5 \sqrt[3] {9}... EQ (5)

From equations 3, 4 and 5 we get:

3 a 5 9 3 3 a 4 3 3 + 48 3 3 96 9 3 = 0... E Q ( 6 ) 3a ^ 5 \sqrt[3] {9} - 3a ^ 4 \sqrt[3] {3} + 48 \sqrt[3] {3} - 96 \sqrt[3] {9} = 0... EQ (6)

Simplifying equation 6 by:

3 3 3 3 \sqrt[3] {3}

we get:

a 5 3 3 a 4 + 16 32 3 3 = 0 a ^ 5 \sqrt[3] {3} - a ^ 4 + 16 - 32 \sqrt[3] {3} = 0
( a 5 32 ) 3 3 ( a 4 16 ) = 0 (a ^ 5 - 32) \sqrt[3] {3} - (a ^ 4 - 16) = 0

Let b = cubic root of 3 and factoring we get :

( a 2 ) ( a 4 + 2 a 3 + 4 a 2 + 8 a + 16 ) ( b ) ( a 2 ) ( a 3 + 2 a 2 + 4 a + 8 ) = 0 (a - 2) (a ^ 4 + 2a ^ 3 + 4a ^ 2 + 8a + 16) (b) - (a - 2) (a ^ 3 + 2a ^ 2 + 4a + 8) = 0

Taking (a – 2) as a common factor we get:

( a 2 ) ( ( a 4 + 2 a 3 + 4 a 2 + 8 a + 16 ) ( b ) ( a 3 + 2 a 2 + 4 a + 8 ) ) = 0 (a - 2) ( (a ^ 4 + 2a ^ 3 + 4a ^ 2 + 8a + 16) (b) - (a ^ 3 + 2a ^ 2 + 4a + 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 + 16 ) ( b ) ( a 3 + 2 a 2 + 4 a + 8 ) = 0 (a ^ 4 + 2a ^ 3 + 4a ^ 2 + 8a + 16) (b) - (a ^ 3 + 2a ^ 2 + 4a + 8) = 0 ( a 4 + ( 2 b 1 ) a 3 + ( 4 b 2 ) a 2 + ( 8 b 4 ) a + ( 16 b 8 ) = 0 (a ^ 4 + (2b - 1) a ^ 3 + (4b - 2) a ^ 2 + (8b - 4) a + (16b - 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 x = 2\sqrt[3] {3}

And

x 3 = 24 x ^ 3 = 24

To calculate the infinite root let’s make it equal to z. we get:

z = 24 + 24 + 24 + . . . 3 3 3 z = \sqrt[3] {24 + \sqrt[3] {24 + \sqrt[3] {24 + ... } } } z = 24 + z 3 z = \sqrt[3] {24 + z}

Cubing both sides:

z 3 = z + 24 z ^ 3 = z + 24 z 3 z 24 = 0 z ^ 3 - z - 24 = 0

Factoring we get:

( z 3 ) ( z 2 + 3 z + 8 ) = 0 (z - 3) (z ^ 2 + 3z + 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 Answer = \boxed {3} .

I think it will be easier start with y = that equation y = \text{that equation} , we can see that y 5 + y 4 = x 5 + x 4 y^5 + y^4 = x^5 + x^4 , so we need to show that y y is a one-to-one function such that y = x y=x is a unique solution. We're left to show that ( x 3 24 ) (x^3-24) is a factor of the equation x = x 5 + 48 3 3 96 9 3 4 x = \sqrt[4] {x ^ 5 + 48 \sqrt[3] {3} - 96 \sqrt[3] {9}} . Thoughts?

Pi Han Goh - 4 years, 10 months ago

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