find the real and complex solutions of the equation x^2-8=0?

Question

possible answers:

a. -2, 2, 2cis(pi/3)
b. 2, 2cis(2pi/3), 2cis(4pi/3)
c. -2, 2cis(pi/3), 2cis(2pi/3)
d. 2cis(pi/3), 2cis(2pi/3), 2cis(4pi/3)

this is straight from the book...?? x^2-8=0 and the possible answer are provided

Answer 1

First, let me state that it is impossible for a second degree function (i.e., the greatest power is 2, as in x² - 3x + 2 = 0) to have THREE solutions... by definition, it can only have a maximum of two solutions.

But anyway...

x² - 8 = 0

Factor and you get:

(x + √[8])(x - √[8]) = 0

Therefore
x = ± √[8]

or
x = ± 2√[2]

If you meant x³ - 8 = 0:

Factor:

(x - 2)(x² + 2x + 4) = 0

Because (x - 2) = 0, we know our first solution is x = 2.

You can't factor x² + 2x + 4, so use the quadratic formula to get the last two solutions.

{-2 ± √[2² - 4(1)(4)]} / 2(1)

{-2 ± √[-12]} / 2

-1 ± √[-3]

x = -1 + √[-3], x = -1 - √[-3]

x = -1 + i √[3], x = -1 - i √[3]

Now, to express this answer using De Moivre's Theorem...

Write z = -1 ± i √[3] in polar form.

|z| = √[a² + b²] =

√[(-1)² + (±√[3])²] =

√[1 + 3] = √[4] =

|z| = 2

Here's the diagram I drew for you:

http://img440.imageshack.us/img440/4784/testad6.png

For the solution x = -1 + i √[3], we can see that z is in the second quadrant. Because

tan α = -√[3],

we know that α = 2π/3.

Therefore, we know that

z = 2 (cos(2π/3) + i sin(2π/3))

z = 2 cis(2π/3)

For the other solution, z = -1 - i √[3]

we can see that z is in the third quadrant. Because

tan α = √[3],

we know that α = 4π/3.

Therefore, we know that

z = 2 (cos(4π/3) + i sin(4π/3))

z = 2 cis(4π/3)


SO, our solutions are x = 2, x = 2 cis(2π/3), and x = 2 cis(4π/3).

This corresponds to answer B.

Answer 2

None of the answers are correct. I think you meant x^3-8 =0, and not x^2-8 = 0. If so, the answer is b because it is the only answer that includes +2 as the real answer.

Answer 3

It has two real solutions 2*sqrt(2), -2*sqrt(2).