What are the solutions to the equation (2x-1)^6=(2x-1)^8 ?

The solutions to the equation (2x-1)^6=(2x-1)^8 are x=0,x=1,x= 1/2

Find the zeros of (2x-1)^6=(2x-1)^8

The zeros of (2x-1)^6=(2x-1)^8 are x=0,x=1,x= 1/2

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(2 x - 1)^{6} = (2 x - 1)^{8}

x = 0, x = 1, x = \frac{1}{2}

Decimal Notation

x = 0, x = 1, x = 0.5

Solution steps

(2 x - 1)^{6} = (2 x - 1)^{8}

Expand

(2 x - 1)^{6} : 64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x + 1

Expand

(2 x - 1)^{8} : 256 x^{8} - 1024 x^{7} + 1792 x^{6} - 1792 x^{5} + 1120 x^{4} - 448 x^{3} + 112 x^{2} - 16 x + 1

64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x + 1 = 256 x^{8} - 1024 x^{7} + 1792 x^{6} - 1792 x^{5} + 1120 x^{4} - 448 x^{3} + 112 x^{2} - 16 x + 1

Switch sides

256 x^{8} - 1024 x^{7} + 1792 x^{6} - 1792 x^{5} + 1120 x^{4} - 448 x^{3} + 112 x^{2} - 16 x + 1 = 64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x + 1

Move

1

to the right side

256 x^{8} - 1024 x^{7} + 1792 x^{6} - 1792 x^{5} + 1120 x^{4} - 448 x^{3} + 112 x^{2} - 16 x = 64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x

Subtract

64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x

from both sides

256 x^{8} - 1024 x^{7} + 1792 x^{6} - 1792 x^{5} + 1120 x^{4} - 448 x^{3} + 112 x^{2} - 16 x - (64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x) = 64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x - (64 x^{6} - 192 x^{5} + 240 x^{4} - 160 x^{3} + 60 x^{2} - 12 x)

Simplify

256 x^{8} - 1024 x^{7} + 1728 x^{6} - 1600 x^{5} + 880 x^{4} - 288 x^{3} + 52 x^{2} - 4 x = 0

Factor

256 x^{8} - 1024 x^{7} + 1728 x^{6} - 1600 x^{5} + 880 x^{4} - 288 x^{3} + 52 x^{2} - 4 x : 4 x (x - 1) (2 x - 1)^{6}

4 x (x - 1) (2 x - 1)^{6} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or 2 x - 1 = 0

Solve

x - 1 = 0 : x = 1

Solve

2 x - 1 = 0 : x = \frac{1}{2}

The solutions are

x = 0, x = 1, x = \frac{1}{2}

C^{8} = \frac{ln ( 7 ) + C}{7}

4 x^{4} + 11 x^{2} - 3 = 0

p^{3} = - 64

42 x^{3} + 30 x^{2} = 21 x + 15

- 3 x^{3} + 2 x^{2} + 8 x = 0

What are the solutions to the equation (2x-1)^6=(2x-1)^8 ?
The solutions to the equation (2x-1)^6=(2x-1)^8 are x=0,x=1,x= 1/2
Find the zeros of (2x-1)^6=(2x-1)^8
The zeros of (2x-1)^6=(2x-1)^8 are x=0,x=1,x= 1/2