# Real and complex solutions

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• Jan 13th 2007, 10:11 PM
mike1
Real and complex solutions
Can someone help me with this one?

Find the real and complex solution of the equation x[squared]-8=0.

Possible answers:
1. -2, 2, 2cis(pi/3)
2. 2, 2cis(2pi/3), 2cis(4pi/3)
3. -2, 2cis(pi/3), 2cis(2pi/3)
4. 2cis(pi/3), 2cis(2pi/3), 2cis(4pi/3)
• Jan 14th 2007, 12:20 AM
CaptainBlack
Quote:

Originally Posted by mike1
Can someone help me with this one?

Find the real and complex solution of the equation x[squared]-8=0.

Possible answers:
1. -2, 2, 2cis(pi/3)
2. 2, 2cis(2pi/3), 2cis(4pi/3)
3. -2, 2cis(pi/3), 2cis(2pi/3)
4. 2cis(pi/3), 2cis(2pi/3), 2cis(4pi/3)

Are you sure that you have typed the question correctly. the only solutions
of $\displaystyle x^2-8=0$, are $\displaystyle x=\pm \sqrt{8}$.

From the list of solutions I would guess you are asked for the solutions of:

$\displaystyle x^3-8=0$,

in which case the answer will be 2, which you can check by cubing each of
the solutions in the lis to check that they come to 8.

RonL
• Jan 14th 2007, 12:36 AM
ticbol
Quote:

Originally Posted by mike1
Can someone help me with this one?

Find the real and complex solution of the equation x[squared]-8=0.

Possible answers:
1. -2, 2, 2cis(pi/3)
2. 2, 2cis(2pi/3), 2cis(4pi/3)
3. -2, 2cis(pi/3), 2cis(2pi/3)
4. 2cis(pi/3), 2cis(2pi/3), 2cis(4pi/3)

If x[squared] -8 = 0
is x^2 -8 = 0
then,
(x+sqrt(8))(x-sqrt(8)) = 0
x = +,-sqrt(8)
x = +,-2sqrt(2) ------------answer.

Another way,
x^2 -8 = 0
x^2 = 8
x = +,-sqrt(8)
.....

The answers are not in the multiple choice.
The answers are all real numbers. No complex.
• Jan 14th 2007, 03:25 AM
mike1
Now I'm really confused. Do you really think my worksheet is wrong?
• Jan 14th 2007, 04:33 AM
CaptainBlack
Quote:

Originally Posted by mike1
Now I'm really confused. Do you really think my worksheet is wrong?

The equation given to solve is a quadratic, the solution sets contain
three solutions, which is a charateristic in the complex domain of the
roots of a cubic.

So yes there is a mistake somewhere.

RonL