What are the solutions to the equation 4x^3-6x^2+1=0 ?

The solutions to the equation 4x^3-6x^2+1=0 are x= 1/2 ,x=(1+sqrt(3))/2 ,x=(1-sqrt(3))/2

Find the zeros of 4x^3-6x^2+1=0

The zeros of 4x^3-6x^2+1=0 are x= 1/2 ,x=(1+sqrt(3))/2 ,x=(1-sqrt(3))/2

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4 x^{3} - 6 x^{2} + 1 = 0

x = \frac{1}{2}, x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

Decimal Notation

x = 0.5, x = 1.36602 \dots, x = - 0.36602 \dots

Solution steps

4 x^{3} - 6 x^{2} + 1 = 0

Factor

4 x^{3} - 6 x^{2} + 1 : (2 x - 1) (2 x^{2} - 2 x - 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

2 x^{2} - 2 x - 1 = 0 : x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

The solutions are

x = \frac{1}{2}, x = \frac{1 + 3}{2}, x = \frac{1 - 3}{2}

x^{4} - 5 x = 0

20 u^{4} - 13 u^{2} + 2 = 0

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

445 = 5 x^{3}

x^{3} + 3 x^{2} - 2 x + 6 = 0

What are the solutions to the equation 4x^3-6x^2+1=0 ?
The solutions to the equation 4x^3-6x^2+1=0 are x= 1/2 ,x=(1+sqrt(3))/2 ,x=(1-sqrt(3))/2
Find the zeros of 4x^3-6x^2+1=0
The zeros of 4x^3-6x^2+1=0 are x= 1/2 ,x=(1+sqrt(3))/2 ,x=(1-sqrt(3))/2