What are the solutions to the equation x^2(x^2-1)=16(x^2-1) ?

The solutions to the equation x^2(x^2-1)=16(x^2-1) are x=-1,x=1,x=-4,x=4

Find the zeros of x^2(x^2-1)=16(x^2-1)

The zeros of x^2(x^2-1)=16(x^2-1) are x=-1,x=1,x=-4,x=4

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x^{2} (x^{2} - 1) = 16 (x^{2} - 1)

x = - 1, x = 1, x = - 4, x = 4

Solution steps

x^{2} (x^{2} - 1) = 16 (x^{2} - 1)

Move

16 (x^{2} - 1)

to the left side

x^{2} (x^{2} - 1) - 16 (x^{2} - 1) = 0

Factor

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

(x + 1) (x - 1) (x + 4) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 1 = 0 or x + 4 = 0 or x - 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = - 1, x = 1, x = - 4, x = 4

x^{3} - 4 x^{2} + 13 x = 0

x^{3} - 5 x + 6 = 2.2

x + x^{2} + 400 x^{3} = 11500

6 x^{4} - x^{3} - 22 x^{2} + 11 x + 6 = 0

4 x^{4} - 349 x^{2} + 2025 = 0

What are the solutions to the equation x^2(x^2-1)=16(x^2-1) ?
The solutions to the equation x^2(x^2-1)=16(x^2-1) are x=-1,x=1,x=-4,x=4
Find the zeros of x^2(x^2-1)=16(x^2-1)
The zeros of x^2(x^2-1)=16(x^2-1) are x=-1,x=1,x=-4,x=4