What are the solutions to the equation (1-x)^5=(2x-1)^5 ?

The solutions to the equation (1-x)^5=(2x-1)^5 are x= 2/3

Find the zeros of (1-x)^5=(2x-1)^5

The zeros of (1-x)^5=(2x-1)^5 are x= 2/3

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(1 - x)^{5} = (2 x - 1)^{5}

x = \frac{2}{3}

Decimal Notation

x = 0.66666 \dots

Solution steps

(1 - x)^{5} = (2 x - 1)^{5}

Expand

(1 - x)^{5} : 1 - 5 x + 10 x^{2} - 10 x^{3} + 5 x^{4} - x^{5}

Expand

(2 x - 1)^{5} : 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1

1 - 5 x + 10 x^{2} - 10 x^{3} + 5 x^{4} - x^{5} = 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1

Subtract

32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1

from both sides

1 - 5 x + 10 x^{2} - 10 x^{3} + 5 x^{4} - x^{5} - (32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1) = 32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1 - (32 x^{5} - 80 x^{4} + 80 x^{3} - 40 x^{2} + 10 x - 1)

Simplify

- 33 x^{5} + 85 x^{4} - 90 x^{3} + 50 x^{2} - 15 x + 2 = 0

Factor

- 33 x^{5} + 85 x^{4} - 90 x^{3} + 50 x^{2} - 15 x + 2 : - (3 x - 2) (11 x^{4} - 21 x^{3} + 16 x^{2} - 6 x + 1)

- (3 x - 2) (11 x^{4} - 21 x^{3} + 16 x^{2} - 6 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 2 = 0 or 11 x^{4} - 21 x^{3} + 16 x^{2} - 6 x + 1 = 0

Solve

3 x - 2 = 0 : x = \frac{2}{3}

Solve

11 x^{4} - 21 x^{3} + 16 x^{2} - 6 x + 1 = 0 :

No Solution for

x \in R

The solution is

x = \frac{2}{3}

a^{5} = 12 - 3 (5)

- x^{3} + 4 x^{2} + 12 x = 0

d^{3} \cdot 1 = 512 \cdot 1

y^{3} + 6 y^{2} - 7 y = 8

x^{4} - 2 x^{2} - 35 = 0

What are the solutions to the equation (1-x)^5=(2x-1)^5 ?
The solutions to the equation (1-x)^5=(2x-1)^5 are x= 2/3
Find the zeros of (1-x)^5=(2x-1)^5
The zeros of (1-x)^5=(2x-1)^5 are x= 2/3