Dnas Posted March 20, 2004 Share Posted March 20, 2004 last samurai is correct.Explanationx^4+x^2+1=x^4+2x^2+1-x^2= (add and subtract x^2)(x^2+1)^2-x^2= ((a+^2=a^2+2ab+b^2)(x^2+x+1)(x^2-x+1) (a^2-b^2=(a+(a-)In case anyone wants to know how to do my other one,A polynomial of degree 4, in the form x^4+ax^3+bx^2+cx+d=0 (a,b, and c are all real numbers), has roots -4+3i and 9+4i. What is the value of a?For a polynomial to have real coeficcients and still have complex roots, each complex root must be paired with its conjugate. So, if two roots are -4+3i and 9+4i, the other two must be -4-3i and 9-4i.In any polynomial in the form,ax^n+bx^(n-1)+cx^(n-2)+dx^(n-3)+ex^(n-4)...-b/a is always the sum of the roots of the equation.So, -a/1=-a= (-4+3i)+(9+4i)+(-4-3i)+(9-4i)=10So, a = -10 Link to comment Share on other sites More sharing options...
CodeOptimist Posted March 20, 2004 Author Share Posted March 20, 2004 Nice explanation, Dnas Link to comment Share on other sites More sharing options...
CodeOptimist Posted March 23, 2004 Author Share Posted March 23, 2004 If last_samurai doesn't ask a new question in a little while, I'll post up a new one Link to comment Share on other sites More sharing options...
akya Posted March 24, 2004 Share Posted March 24, 2004 Akya - Oh well, if nobody else is going to guess:f(x,y) = x^2 + 4xy + sin y + y^2(df/dx)y = 2x + 4y(df/dy)x = 4x + cos y + 2y yep got it right...^^ sorry I didn't answer immediatly how about this :Find a polar equation for the circle x^2+(y-3)^2=9 Link to comment Share on other sites More sharing options...
King Tutankhamun Posted March 24, 2004 Share Posted March 24, 2004 Hey that is so cool, we just learned that equation today!The radius is 3, the center is (0, 3) I think.Is that what you are asking for? Link to comment Share on other sites More sharing options...
MarkT Posted March 24, 2004 Share Posted March 24, 2004 Polar-cartesian substitutionsx = r * cos( t )y = r * sin( t )x ^ 2 + ( y - 3 ) ^ 2 = 9r^2 * cos(t)^2 + r^2 * sin(t)^2 - 6 * r * sin(t) + 9 = 9r^2 * ( cos(t)^2 + sin(t)^2 ) - 6r*sin(t) = 0r^2 - 6r*sin(t) = 0r = 6 * sin(t) for 0 <= t < pi/2 Link to comment Share on other sites More sharing options...
King Tutankhamun Posted March 24, 2004 Share Posted March 24, 2004 Humm, I guess that I thought it was too simple. Link to comment Share on other sites More sharing options...
CodeOptimist Posted March 24, 2004 Author Share Posted March 24, 2004 Humm, I guess that I thought it was too simple. Hehe, I don't blame you... Did Mark get the correct answer? Link to comment Share on other sites More sharing options...
akya Posted March 25, 2004 Share Posted March 25, 2004 yep his turn to ask a Q I guess Link to comment Share on other sites More sharing options...
CodeOptimist Posted March 28, 2004 Author Share Posted March 28, 2004 Tick tick... Mark, where art thou? Link to comment Share on other sites More sharing options...
King Tutankhamun Posted March 28, 2004 Share Posted March 28, 2004 Anybody can feel free to ask the next one if ever someone does not relpy within a while. Link to comment Share on other sites More sharing options...
Curufinwe Posted March 28, 2004 Share Posted March 28, 2004 (okay, factorize :A= (-25 + 9x²) - (25 + 9x² - 30x) Link to comment Share on other sites More sharing options...
Dnas Posted March 28, 2004 Share Posted March 28, 2004 30x-50 = 10(3x-5) Link to comment Share on other sites More sharing options...
chichigrande Posted March 28, 2004 Share Posted March 28, 2004 A= (-25 + 9x²) - (25 + 9x² - 30x)-30x - -50 -> -30x + 50 -> -30x = -50 -> x = 5/3Did I do it right? Link to comment Share on other sites More sharing options...
last samurai Posted March 31, 2004 Share Posted March 31, 2004 A=10(3X-5) Link to comment Share on other sites More sharing options...
Argalius Posted April 4, 2004 Share Posted April 4, 2004 I got this:A= (-25 + 9x²) - (25 + 9x² - 30x)A= 9x² - 50 + 9x² - 30xA= - 50 - 30xA= 10(- 5 - 3x)A= 10(- 3x - 5) or just - 30x - 50 Link to comment Share on other sites More sharing options...
Curufinwe Posted April 4, 2004 Share Posted April 4, 2004 (These are all correct ...A = 10(3x-5)Dnas can ask a new question. Link to comment Share on other sites More sharing options...
Argalius Posted April 4, 2004 Share Posted April 4, 2004 What did I do wrong? Link to comment Share on other sites More sharing options...
Dnas Posted April 4, 2004 Share Posted April 4, 2004 I've attached it as an image.Simplify.Note: Please show some sort of work to prove that you didn't use a calculator.If this is too hard, I'll post a different one. Link to comment Share on other sites More sharing options...
last samurai Posted April 4, 2004 Share Posted April 4, 2004 that is way nuts, too hard Link to comment Share on other sites More sharing options...
akya Posted April 4, 2004 Share Posted April 4, 2004 dunno if I got it Link to comment Share on other sites More sharing options...
Dnas Posted April 4, 2004 Share Posted April 4, 2004 Nope.I'll give a few hints, then if needed, I'll change it.First hint:Try cubing it. Link to comment Share on other sites More sharing options...
Mystic-Al-Bob Posted April 6, 2004 Share Posted April 6, 2004 @ akja: nice try but you can't just connect the rootsI also think cubing is a good idea Link to comment Share on other sites More sharing options...
Argalius Posted April 6, 2004 Share Posted April 6, 2004 I have no idea what cubing is but this is what I think:2³√2, can't think of anything else. Link to comment Share on other sites More sharing options...
MarkT Posted April 7, 2004 Share Posted April 7, 2004 Deep breath... and:A = (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3)A^3 = (2+sqrt(5)) + 3.((2+sqrt(5))(9-4.sqrt(5)))^(1/3) + 3.((2-sqrt(5))(9+4.sqrt(5)))^(1/3) + (2-sqrt(5))= 4 - 3.[ (2+sqrt(5))^(1/3) + (2-sqrt(5))^(1/3) ]= 4 - 3.AA^3 + 3.A - 4 = 0(A-1)(A^2+A+4) = 0A = 1. I'm not going to bother with the imaginary roots.Good enough? Link to comment Share on other sites More sharing options...
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