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

The solutions to the equation x^2(x^2-4)=x^4(x^2-4) are x=0,x=-2,x=2,x=-1,x=1

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

The zeros of x^2(x^2-4)=x^4(x^2-4) are x=0,x=-2,x=2,x=-1,x=1

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x^{2} (x^{2} - 4) = x^{4} (x^{2} - 4)

x = 0, x = - 2, x = 2, x = - 1, x = 1

Solution steps

x^{2} (x^{2} - 4) = x^{4} (x^{2} - 4)

Move

x^{4} (x^{2} - 4)

to the left side

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 2 = 0 or x - 2 = 0 or x + 1 = 0 or x - 1 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

x = 0, x = - 2, x = 2, x = - 1, x = 1

x^{4} - 5 x^{3} + 4 = 0

x^{3} - x^{2} - x - 25 = 470

- x^{3} - x^{2} + 8 x + 12 = 0

2 a^{2} (a^{2} + 2) = 27 + 7 a^{2}

2 x - 53 = 2 + x^{15}

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