Hi. Badly need help here. What's the solution to this? x^4 + 64= 0?

Answer 1

Well, we have x⁴ = -64, so we need to find the fourth roots of 64. We have, by Euler's formula:

-64 = 64e^(iπ+2iπk), where k is any integer

so the fourth root would be:

∜64 e^(iπ/4 + iπk/2).

64 = 2^6, so ∜64 = 2^(3/2) = 2√2, and there are four distinct solutions for e^(iπ/4 + iπk/2) -- namely, e^(iπ/4), e^(-iπ/4), e^(3iπ/4), and e^(-3iπ/4). We know that:

e^(iπ/4) = cos (π/4) + i sin (π/4) = 1/√2 + i/√2

Therefore our first solution is:

x=2√2 * (1/√2 + i/√2) = 2+2i.

Proceeding in this fashion, the other solutions are x=2-2i, x=-2+2i, and x=-2-2i.

Check: calculating (2+2i)⁴, we find that

(2+2i)⁴ = ((2+2i)²)² = (4 + 2*4i - 4)² = (8i)² = -64 ✓
(2-2i)⁴ = ((2-2i)²)² = (4 - 2*4i - 4)² = (-8i)² = -64 ✓
(-2+2i)⁴ = ((-2+2i)²)² = (4 - 2*4i - 4)² = (-8i)² = -64 ✓
(-2-2i)⁴ = ((-2-2i)²)² = (4 + 2*4i - 4)² = (8i)² = -64 ✓

So our solutions are correct, and we are done.

Answer 2

Well for a start, it's a complex number because any RATIONAL number, raised to an even power, becomes positive.
Also, 2^4 = 16 and 3^4 = 81
so it will be between (0+2i) and (0+3i)

2√2 = √8
(√8)^2 = 8
(√8)^4 = 64

so try: 0 +/- (2√2)i
--- Edit ---
Pascal (below) has the better solution.
Mine becomes a negative (rational) number when squared, which becomes a POSITIVE number at fourth power.

Answer 3

if complex number is not included, there will be no solution to
x^4 + 64 = 0 as the minimum value of x^4 will be zero, x^4 + 64 will be at least 64.

Answer 4

x^4=-64
x=4throot(-64)
x=-2.828427i

(Take the "i" out manually so you can use your calculator on it.)

Ooops, this goes beyond "i"... I think the problem is undefined.

We will have to make up hyper-hyper complex numbers with "k" = 4throot(-1) to solve this.

Answer 5

2 sqrt(2i)