thedarkshade Posted January 9, 2008 Share Posted January 9, 2008 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] Link to comment Share on other sites More sharing options...
NeonBlack Posted January 9, 2008 Share Posted January 9, 2008 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. Link to comment Share on other sites More sharing options...
thedarkshade Posted January 9, 2008 Share Posted January 9, 2008 OK, I must have "overunderstood". Imaginary numbers are within complex number! Link to comment Share on other sites More sharing options...
TimbaLanD Posted January 9, 2008 Author Share Posted January 9, 2008 So, x= WHAT? Link to comment Share on other sites More sharing options...
thedarkshade Posted January 9, 2008 Share Posted January 9, 2008 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... Link to comment Share on other sites More sharing options...
NeonBlack Posted January 9, 2008 Share Posted January 9, 2008 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? Link to comment Share on other sites More sharing options...
thedarkshade Posted January 9, 2008 Share Posted January 9, 2008 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] Link to comment Share on other sites More sharing options...
ydoaPs Posted January 9, 2008 Share Posted January 9, 2008 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. Link to comment Share on other sites More sharing options...
5614 Posted January 9, 2008 Share Posted January 9, 2008 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? Link to comment Share on other sites More sharing options...
the tree Posted January 9, 2008 Share Posted January 9, 2008 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]). Link to comment Share on other sites More sharing options...
thedarkshade Posted January 9, 2008 Share Posted January 9, 2008 yourdadonapogos: That confusing because + and - should be one on top of other, and I don't know the LaTeX procedure to do that... Link to comment Share on other sites More sharing options...
iNow Posted January 9, 2008 Share Posted January 9, 2008 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] Link to comment Share on other sites More sharing options...
thedarkshade Posted January 9, 2008 Share Posted January 9, 2008 thnx iNow Link to comment Share on other sites More sharing options...
iNow Posted January 9, 2008 Share Posted January 9, 2008 You're welcome, shade. Cheers. Link to comment Share on other sites More sharing options...
TimbaLanD Posted January 10, 2008 Author Share Posted January 10, 2008 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] Link to comment Share on other sites More sharing options...
ydoaPs Posted January 10, 2008 Share Posted January 10, 2008 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? Link to comment Share on other sites More sharing options...
TimbaLanD Posted January 10, 2008 Author Share Posted January 10, 2008 [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] Link to comment Share on other sites More sharing options...
ydoaPs Posted January 10, 2008 Share Posted January 10, 2008 Oh, so you were just splitting up the polynomial under the radical. Now, keep going. Link to comment Share on other sites More sharing options...
TimbaLanD Posted January 10, 2008 Author Share Posted January 10, 2008 [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? Link to comment Share on other sites More sharing options...
ydoaPs Posted January 10, 2008 Share Posted January 10, 2008 Is 5-2+1 equal to 2 or 4? Link to comment Share on other sites More sharing options...
TimbaLanD Posted January 10, 2008 Author Share Posted January 10, 2008 4, where did i go wrong? Link to comment Share on other sites More sharing options...
ydoaPs Posted January 10, 2008 Share Posted January 10, 2008 You are supposed to do addition and subtraction in order from left to right. Link to comment Share on other sites More sharing options...
swansont Posted January 10, 2008 Share Posted January 10, 2008 [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) Link to comment Share on other sites More sharing options...
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