Thread: 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)

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 , are .

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

,

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

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.

Now I'm really confused. Do you really think my worksheet is wrong?

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