What are the solutions to the equation x^3-x= 1/4 x^2-1/4 ?

The solutions to the equation x^3-x= 1/4 x^2-1/4 are x= 1/4 ,x=-1,x=1

Find the zeros of x^3-x= 1/4 x^2-1/4

The zeros of x^3-x= 1/4 x^2-1/4 are x= 1/4 ,x=-1,x=1

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x^{3} - x = \frac{1}{4} x^{2} - \frac{1}{4}

x = \frac{1}{4}, x = - 1, x = 1

Decimal Notation

x = 0.25, x = - 1, x = 1

Solution steps

x^{3} - x = \frac{1}{4} x^{2} - \frac{1}{4}

Multiply both sides by

4

4 x^{3} - 4 x = x^{2} - 1

Move

1

to the left side

4 x^{3} - 4 x + 1 = x^{2}

Move

x^{2}

to the left side

4 x^{3} - 4 x + 1 - x^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

4 x^{3} - x^{2} - 4 x + 1 = 0

Factor

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

(4 x - 1) (x + 1) (x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 x - 1 = 0 or x + 1 = 0 or x - 1 = 0

Solve

4 x - 1 = 0 : x = \frac{1}{4}

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

x = \frac{1}{4}, x = - 1, x = 1

x^{3} + 3 x^{2} + 3 x - 7 = 0

8 x^{1} + x^{2} + 2 x^{3} + x^{4} = 8

x, y^{2} + 1 = x^{3}

x^{4} - 4 x^{3} + 8 x^{2} - 8 x + 4 = 0

x^{4} - x^{2} - 1 = 0

What are the solutions to the equation x^3-x= 1/4 x^2-1/4 ?
The solutions to the equation x^3-x= 1/4 x^2-1/4 are x= 1/4 ,x=-1,x=1
Find the zeros of x^3-x= 1/4 x^2-1/4
The zeros of x^3-x= 1/4 x^2-1/4 are x= 1/4 ,x=-1,x=1