What are the solutions to the equation 2b^4-32=0 ?

The solutions to the equation 2b^4-32=0 are b=2,b=-2,b=2i,b=-2i

Find the zeros of 2b^4-32=0

The zeros of 2b^4-32=0 are b=2,b=-2,b=2i,b=-2i

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2 b^{4} - 32 = 0

b = 2, b = - 2, b = 2 i, b = - 2 i

Solution steps

2 b^{4} - 32 = 0

Move

32

to the right side

2 b^{4} = 32

Divide both sides by

2

b^{4} = 16

Rewrite the equation with

u = b^{2}

and

u^{2} = b^{4}

u^{2} = 16

Solve

u^{2} = 16 : u = 16, u = - 16

u = 16, u = - 16

Substitute back

u = b^{2},

solve for

b

Solve

b^{2} = 16 : b = 2, b = - 2

Solve

b^{2} = - 16 : b = 2 i, b = - 2 i

The solutions are

b = 2, b = - 2, b = 2 i, b = - 2 i

x^{2} = 196 x^{4}

114 x^{4} - 2 4^{2} + 1 = 0

- a^{3} + 7 = - 20

82853.3352 = \frac{4}{3} π r^{3}

x^{3} + 3 x^{2} + 3 x - 8059 = 0

What are the solutions to the equation 2b^4-32=0 ?
The solutions to the equation 2b^4-32=0 are b=2,b=-2,b=2i,b=-2i
Find the zeros of 2b^4-32=0
The zeros of 2b^4-32=0 are b=2,b=-2,b=2i,b=-2i