solve (x-4)-5(x-4)^1/4 = 6?

(x-4)-5(x-4)^(1/4) = 6

let (x - 4)^(1/4) = u

u^4 - 5u - 6 = 0

(u^2 + u + 3)(u^2 - u - 2) = 0

u^2 + u + 3 = 0 => reject (complex roots)

u^2 - u - 2 = 0

(u + 1)(u - 2) = 0

u = -1 , 2

(x - 4)^(1/4) = -1 => reject(complex roots)

(x - 4)^(1/4) = 2

x - 4 = 2^4 = 16

x = 20

Use a substitution of a complicated expression to something simple to make things easier to manage. Later, we will reverse the substitution, after we have a much simpler expression.

Let u=(x-4)^(1/4)

Therefore, (x-4) = u^4

u^4 - 5u = 6

u^4 - 5u - 6 = 0

Using factor theorem, the rational solutions only can happen when u = {-6, -3, -2, -1, 1, 2, 3, 6}. Substitute each value in for u; when the result is 0, you have a factor.

(-1)^4 - 5(-1) - 6 = 1 + 5 - 6 = 0

Therefore u+1 is a factor of this quartic expression. Factor out u+1.

(u^4) - 5u - 6 = (u+1)(u^3 - u^2 + u - 6)

(2)^3 - (2)^2 + (2) - 6 = 8 - 4 + 2 - 6 = 0

Therefore u-2 is also a factor.

(u+1)(u^3 - u^2 + u - 6) = (u+1)(u-2)(u^2 + u + 3)

For the factors of the quadratic, use the quadratic solution.

u = [-b +- sqrt(b^2 - 4ac) ] / (2a)

u = [ -1 +- sqrt(1^2 - 4[1][3]) ] / 2(1)

u = [ -1 +- sqrt(1 - 12) ] / 2

u = [ -1 +- sqrt(-11) ] / 2

We can not take the square root of -11, so there are no other real factors. So u+1=0 or u-2 = 0

u+1 = 0 OR u - 2 = 0

u= -1 OR u = 2

Now reverse the earlier substitution:

(x-4)^(1/4) = -1 OR (x-4)^(1/4) = 2

Raise both sides to the power of 4.

x-4 = 1 OR x-4 = 16

x = 5 OR x = 20

Test each value in the original equation.

LS=(5-4) - 5(5-4)^(1/4) = 1 - 5(1) = -4

This is not an actual answer.

LS = (20-4) - 5(20-4)^(1/4) = 16 - 5(16)^(1/4) = 16 - 5(2) = 16 - 10 = 6

So x = 20 is the valid answer.

Let (x-4)^¼ = A

Then (x-4) = A⁴

A⁴ - 5A - 6 = 0

The rational roots theorem says that ±{1,2,3,6} are the only possible rational roots to the equation. By inspection A = -1 is a root. Synthetic division:

-1 | 1. . . 0. . . 0. . . -5. . . -6

_______-1___1___-1___6__

. . . 1. . . -1. . . 1. . . -6. . . 0, confirmed, x = -1 is a root

The bottom row represents one degree less than the 4th degree we started with

x³ - x² + x - 6 = 0

2 | 1. . .-1. . .1. . . -6

_______2__2____6_

. . .1. . . 1. . . 3. . . 0, thus x = 2 is a root.

Solve the resulting quadratic x²+x+3 = 0 → x = (-1± i√11)/2

Now recall our original substitution A = (x-4)^¼

➥ -1 = (x-4)^¼ has no solution because a 4th root is not negative

➥ 2 = (x-4)^¼ → 16 = x-4 → x = 20

➥ (-1± i√11)/2 = (x-4)^(1/4) yields even more complicated imaginary solutions, so you can take the x=20 and run

(x - 4) - 5(x - 4)^1/4 = 6

5(x - 4)^1/4 = (x - 4) - 6

(x - 4)^1/4 = (x - 10)/5

(x - 4) = [(x - 10)/5]^4

625(x - 4) = (x^2 - 20x + 100)^2

625x - 2500 = x^4 - 40x^3 + 600x^2 - 4,000x + 10,000

x^4 - 40x^3 + 600x^2 - 4,625x + 12,500 = 0

(x - 20)(x - 5)(x^2 - 15x + 125) = 0

Real solutions:

x = 5

x = 20

Complex solutions:

x = 5(3 - i sqrt 11)/2

x = 5(3 + i sqrt 11)/2