What are the solutions to the equation 2x(4x^2-4x+1)=x(8x)+4(2x) ?

The solutions to the equation 2x(4x^2-4x+1)=x(8x)+4(2x) are x=0,x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2

Find the zeros of 2x(4x^2-4x+1)=x(8x)+4(2x)

The zeros of 2x(4x^2-4x+1)=x(8x)+4(2x) are x=0,x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x (4 x^{2} - 4 x + 1) = x (8 x) + 4 (2 x)

x = 0, x = \frac{2 + 7}{2}, x = \frac{2 - 7}{2}

Decimal Notation

x = 0, x = 2.32287 \dots, x = - 0.32287 \dots

Solution steps

2 x (4 x^{2} - 4 x + 1) = x (8 x) + 4 (2 x)

Expand

2 x (4 x^{2} - 4 x + 1) : 8 x^{3} - 8 x^{2} + 2 x

Expand

x (8 x) + 4 (2 x) : 8 x^{2} + 8 x

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

Move

8 x

to the left side

8 x^{3} - 8 x^{2} - 6 x = 8 x^{2}

Move

8 x^{2}

to the left side

8 x^{3} - 16 x^{2} - 6 x = 0

Factor

8 x^{3} - 16 x^{2} - 6 x : 2 x (4 x^{2} - 8 x - 3)

2 x (4 x^{2} - 8 x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

4 x^{2} - 8 x - 3 = 0 : x = \frac{2 + 7}{2}, x = \frac{2 - 7}{2}

The solutions are

x = 0, x = \frac{2 + 7}{2}, x = \frac{2 - 7}{2}

18 x^{3} - 27 x^{2} - 50 x + 75 = 0

x^{4} + 4 x^{2} = 0

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

x^{3} - 4 x^{2} + 20 x - 77 = 0

(x + 10)^{5} = 70

What are the solutions to the equation 2x(4x^2-4x+1)=x(8x)+4(2x) ?
The solutions to the equation 2x(4x^2-4x+1)=x(8x)+4(2x) are x=0,x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2
Find the zeros of 2x(4x^2-4x+1)=x(8x)+4(2x)
The zeros of 2x(4x^2-4x+1)=x(8x)+4(2x) are x=0,x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2