What are the solutions to the equation 4a^4=8a^3-8a^2 ?

The solutions to the equation 4a^4=8a^3-8a^2 are a=0,a=1+i,a=1-i

Find the zeros of 4a^4=8a^3-8a^2

The zeros of 4a^4=8a^3-8a^2 are a=0,a=1+i,a=1-i

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4 a^{4} = 8 a^{3} - 8 a^{2}

a = 0, a = 1 + i, a = 1 - i

Solution steps

4 a^{4} = 8 a^{3} - 8 a^{2}

Move

8 a^{2}

to the left side

4 a^{4} + 8 a^{2} = 8 a^{3}

Move

8 a^{3}

to the left side

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

Factor

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

4 a^{2} (a^{2} - 2 a + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

a = 0 or a^{2} - 2 a + 2 = 0

Solve

a^{2} - 2 a + 2 = 0 : a = 1 + i, a = 1 - i

The solutions are

a = 0, a = 1 + i, a = 1 - i

x^{3} - 12 x^{2} + 49 x - 58 = 0

8 = b^{3}

20 x^{3} - 90 x = 0

48.46 x^{3} - 12.6 x + 0.005 = 0

x, x^{5} + x^{3} + x = 3

What are the solutions to the equation 4a^4=8a^3-8a^2 ?
The solutions to the equation 4a^4=8a^3-8a^2 are a=0,a=1+i,a=1-i
Find the zeros of 4a^4=8a^3-8a^2
The zeros of 4a^4=8a^3-8a^2 are a=0,a=1+i,a=1-i