What are the solutions to the equation (x^2+1)(x^2+4)=0 ?

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

Find the zeros of (x^2+1)(x^2+4)=0

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

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(x^{2} + 1) (x^{2} + 4) = 0

x = i, x = - i, x = 2 i, x = - 2 i

Solution steps

(x^{2} + 1) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x^{2} + 1 = 0 or x^{2} + 4 = 0

Solve

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

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

x = i, x = - i, x = 2 i, x = - 2 i

n^{3} - 2 n^{2} + 15 n - 500 = 0

1 x^{4} + 21 x^{2} - 100 = 0

t^{3} = 216

18 q^{3} - 8 q = 0

0 = x^{3} + 4 x^{2} - 9 x - 36

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