What are the solutions to the equation 16x^4+12x^2-4x=0 ?

The solutions to the equation 16x^4+12x^2-4x=0 are x=0,x\approx 0.29803…

Find the zeros of 16x^4+12x^2-4x=0

The zeros of 16x^4+12x^2-4x=0 are x=0,x\approx 0.29803…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 0, x \approx 0.29803 \dots

Solution steps

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

Factor

16 x^{4} + 12 x^{2} - 4 x : 4 x (4 x^{3} + 3 x - 1)

4 x (4 x^{3} + 3 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

4 x^{3} + 3 x - 1 = 0 : x \approx 0.29803 \dots

The solutions are

x = 0, x \approx 0.29803 \dots

3 x^{3} - 24 = 0

x^{4} - 4 x^{2} + 5 = 0

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

12 x^{3} - 48 x^{2} + 36 x = 0

0 = 5 x^{4} - 6 x + 1

What are the solutions to the equation 16x^4+12x^2-4x=0 ?
The solutions to the equation 16x^4+12x^2-4x=0 are x=0,x\approx 0.29803…
Find the zeros of 16x^4+12x^2-4x=0
The zeros of 16x^4+12x^2-4x=0 are x=0,x\approx 0.29803…