No, it'll be complex (have both real and imaginary parts).

Imaginary and complex numbers are the same thing

 

[math]\sqrt{-1}=i[/math]

 

[math]i^2=-1[/math]

 

 

[math]i^3=-i[/math]

 

[math]i^4=1[/math]

Imaginary and complex numbers are not the same. "Imaginary" refers to a multiple of i and "complex" refers to a combination of imaginary and real numbers. Complex numbers can be written in the form

[math]\alpha i + \beta [/math] ,

where alpha and beta are real.

OK, I must have "overunderstood". Imaginary numbers are within complex number!

So, x= WHAT?

So, x= WHAT?

I am kinda lost now in this thread... Why don't you just tell me the equation and I'll to give you the answer...

darkshade: yes, the complex set of numbers is the most general. Imaginary numbers are complex and so are reals. Every (finite) number is complex.

 

TL: I know you must be getting frustrated by now, but we cannot give you the answer. Which part is giving you trouble? Do you know how to find the roots of a general quadratic?

TL: If you're stuck only to quadratic equation then maybe this will help:

 

[math]X_{1/2}=\frac{-b +- \sqrt{b^2 - 4ac}}{2a}[/math]

TL: If you're stuck only to quadratic equation then maybe this will help:

 

[math]X_{1/2}=\frac{-b +- \sqrt{b^2 - 4ac}}{2a}[/math]

 

Your LaTeX is somewhat confusing for those who don't know the formula. You should have split it up and defined a, b, and c.

 

For a quadratic equation(like what the equation in the OP will reduce to), you can get the roots by [math]X_{1/2}=\frac{-b + \sqrt{b^2 - 4ac}}{2a}[/math] and [math]X_{1/2}=\frac{-b - \sqrt{b^2 - 4ac}}{2a}[/math] where the equation is ax²+bx+c=0. Remember that BOTH are roots and they may or may not be imaginary.

For a quadratic of form [math]ax^2 + bx + c[/math] the two roots can be found using the formula [math]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math] (first root from using +, 2nd from -).

 

So just take your equation and stick in the values of a,b,c into that equation.

 

As for the complex part, if you get, for example [math]\sqrt{-100} = \sqrt{-1 * 100} = \sqrt{-1} * \sqrt{100} = 10i[/math]

NB: sqrt(-1) = i

 

I'd stick in the values, but thedarkshade and the tree got slightly different equations, and I haven't checked who is correct. Either way they both give two complex roots to your original equation.

 

 

[edit] in LaTeX should I use \i or just i for an imaginary number? Because \i ([math]\i[/math]) doesn't really look like a good i, whereas using i ([math]i[/math]) is a lot easier to read, but then what's \i for?

edit this post is kind of redundant, feel free to ignore it.

 

Imaginary and complex numbers are the same thing.

No. A complex number has both real an imaginary parts. That is [math]\mathbb{C} = \left \{ a + b i | a , b \in \mathbb{R} \right \}[/math], most complex numbers, are obviously not just imaginary.

 

And even though I lost track of weather we were supposed to be giving the final answer or not, I (and Maple) got [math]-\frac{18}{5}\pm\frac{1}{15} i \sqrt{1119}[/math] (approximately [math]-3.6\pm 2.23 i[/math]).

yourdadonapogos: That confusing because + and - should be one on top of other, and I don't know the LaTeX procedure to do that...

yourdadonapogos: That confusing because + and - should be one on top of other, and I don't know the LaTeX procedure to do that...

 

[math]\pm[/math]

 

[math]\pm[/math]

 

thnx iNow

You're welcome, shade. Cheers.

This is what I got:

 

[math]

x^{3} + 9x^{2} + 27x + 27

[/math]

 

[math]

x^{3} + 3x^{2} + 3x + 1

[/math]

 

[math]

x^{3} - 24x^{2} + 192x - 512

[/math]

This is what I got:

 

[math]

x^{3} + 9x^{2} + 27x + 27

[/math]

 

[math]

x^{3} + 3x^{2} + 3x + 1

[/math]

 

[math]

x^{3} - 24x^{2} + 192x - 512

[/math]

 

How? You seem to have lost your = sign. Could you please show all work?

[math]

(x+3)^3=x^3+3x^2+6x^2+18x+9x+27=x^3+9x^2+27x+27

[/math]

[math]

(x+1)^3=x^3+x^2+2x^2+2x+x+1=x^3+3x^2+3x+1

[/math]

[math]

(x-8)^3=x^3-8x^2-162x^2+128x+64x-5121=x^3-24x^2+192x-512

[/math]

Oh, so you were just splitting up the polynomial under the radical. Now, keep going.

[math]

(x+1)^3+(x-8)^3= 2x^3-21x^2+195x-511

[/math]

[math]

(x+3)^3-(2x^3-21x^2+195x-511)=2x^3-30x^2+168x-538

[/math]

 

Does this look ok?

Is 5-2+1 equal to 2 or 4?

4, where did i go wrong?

You are supposed to do addition and subtraction in order from left to right.

[math]

(x+3)^3-(2x^3-21x^2+195x-511)=2x^3-30x^2+168x-538

[/math]

 

Does this look ok?

 

You've got some sign issues, just from a very brief inspection

 

The first parentheses will give you an x³, and the second has -2x³ (because of the subtraction), so the answer can't lead with 2x³

Also, the first parentheses will only have positive terms in it, and you're subtracting -21x², so you shouldn't end up with a negative (-30x²)