Guest Posted June 11, 2011 Report Share Posted June 11, 2011 3x^2 - 6x + 4 = 0 Solver gets a cookie. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 There is no solution for x Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 There is no solution for x No possible solution whatsoever? Quote Link to comment Share on other sites More sharing options...
0 Anon26 Posted June 11, 2011 Report Share Posted June 11, 2011 1,1 I will PM you my address for the cookie if it is right, I want a real cookie k? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 1,1 I will PM you my address for the cookie if it is right, I want a real cookie k? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 1,1 I will PM you my address for the cookie if it is right, I want a real cookie k? I think the answers are 1,2. Steps? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 I do not think there is any possible solution. x=1 is the equation miniumum, returning a value of 1, not 0. Any larger or smaller x will return a value larger than 1. A graph of this equation never touches zero for any x. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 X = three plus or minus i times the square root of 3 all divided by 3 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 As stated above, the equation has no solution among the real numbers. If you Apply Baskhara, you get this: x = (-b±√∆)/2a or x = (6±√∆)/6 Where ∆ = b² - 4ac or ∆ = 6² -4.3.4 = 36 - 48 = -12 Since ∆ < 0, it's square root is not a real number. But if we let √(-12) = √(-1).√12, and √(-1)=i, where i is the imaginary unit, we get: x = (6±√(-1).√12)/6 x = (6±i√12)/6 x = (6±i√(2²3))/6 x = (6±i2√3)/6 x = (3±i√3)/3 As stated in the previous post. So the 2 solutions are x1=(3±i√3)/3 and x2=(3±i√3)/3 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 C'mon kids! This is basic 9th grade algebra! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 The answer lies within the set of complex numbers. 1+(31/2i)/3 or 1-(31/2i)/3 Quote Link to comment Share on other sites More sharing options...
0 Anon26 Posted June 11, 2011 Report Share Posted June 11, 2011 oops My answer above (1,1) is the vertex of the graph, Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 Hints to solve for x: Use quadratic formula [-b +/- √(b^2-4ac)] / 2a a,b, and c come from ax^2+bx+c = 0 1, -1 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 (edited) Hints to solve for x: Use quadratic formula [-b +/- √(b^2-4ac)] / 2a a,b, and c come from ax^2+bx+c = 0 Right idea 1, -1 Answers are not right as neither satisfy the quadratic. Edited June 11, 2011 by jpf Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 Close as I can get -0.5275252316519470000 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 1+-(sqrt(3)/3)i where i equals the square root of -1 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 11, 2011 Report Share Posted June 11, 2011 archlordbr came close, but then jpf got it first Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 12, 2011 Report Share Posted June 12, 2011 Two of the possible ways are as such: x = 1 ± ⅓∙i∙√3 x1 = 1 - ⅓∙i∙√3 x2 = 1 + ⅓∙i∙√3 archlordbr was the first to present a correct answer, though he did mistakenly keep the plus-minus sign ( ± ) for both roots when he meant plus ( + ) for one and minus ( - ) for the other. Nonetheless, the answer was still correct. Will his cookie be chocolate chip? Quote Link to comment Share on other sites More sharing options...
Question
Guest
3x^2 - 6x + 4 = 0
Solver gets a cookie.
Link to comment
Share on other sites
17 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.