 mersenneforum.org Cuberoot Equation
 Register FAQ Search Today's Posts Mark Forums Read 2003-07-01, 11:50 #1 koal   Nov 2002 Vienna, Austria 4110 Posts Cuberoot Equation :? no cube root sign on the web :( The following equation CUBEROOT (x + 15) - CUBEROOT (x - 11) = 2 has two (obvious) solutions: x1 = 12, x2 = -16 because CR(27)-CR(1) = 3-1 = 2 and CR(-1)-CR(-27) = -1+3 = 2 But most people (teachers inclusive) have problems to solve it. Who of you can find the tricky way get the results above? 8)   2003-07-01, 12:32 #2 xilman Bamboozled!   "𒉺𒌌𒇷𒆷𒀭" May 2003 Down not across 101100001101002 Posts I don't know if this is "tricky", but its easy enough. Write the equation as (x+15)^(1/3) = 2 +(x-11)^1/3) Write y=(x-11)^(1/3) and cube both sides to get x+15 = 8 + 12y + 6y^2 + x-11 or 0 = -3 + 2y + y^2 where I've subtracted x+15 from both sides and cleared a common factor of 6. The quadratic in y is easily seen to factor as (y+3)(y-1) so y = -3 or y=1. Hence, (x-11)^(1/3) = 1 or (x-11)^(1/3) = -3 Cubing, gives x-11 = 1 or x-11 = -27 Hence, x = 12 or -16 as you said. What was hard about that? Paul   2003-07-01, 15:18 #3 hyh1048576   Jun 2003 10000002 Posts Is this tricky enough? Denote a=(x+15)^(1/3),b=(x-11)^(1/3) So a-b=2----------------------------(1) a^3-b^3=26--------------------(2) (2)/(1): a^2+ab+b^2=13-------(3) (3)-(1)^2: 3ab=9 So ab=3 and a-b=2 a=-1 or a=3 x+15=-1 or 27 x=-16 or 12   2003-07-03, 11:58 #4 koal   Nov 2002 Vienna, Austria 41 Posts Correct and both ways tricky, but not the tricky way I ment. I saw students trying to substitute one cube root, saying y = x-11 <=> x = y+11 then getting 1) CR(Y+26) - CR(Y) = 2 cubing both sides gives 2) Y+26 - 3*CR((Y+26)^2*Y) + 3*CR((Y+26)*Y^2) - Y = 8 or 3) 18 - 3*CR((Y+26)*Y) * (CR(Y+26) - CR(Y)) = 0 So they were lost in the wilderness of cube roots and giving up. But what happens, if one substitutes both cube roots with the arithmetic middle of them? y = (x+15+x-11)/2 y = x+2 <=> x =y-2 which gives the equation 1) CR(Y+13) - CR(Y-13) = 2 now it's obvious that (see above) 3) 18 - 3*CR((Y+26)*Y) * (CR(Y+26) - CR(Y)) = 0 becomes 3) 18 - 3*CR((Y+13)*(Y-13)) * (CR(Y+13) - CR(Y-13)) = 0 due to 1) we kann substitute CR(Y+13) - CR(Y-13) = 2 4) 18 - 3*CR((Y+13)*(Y-13)*2 = 0 ... 5) 3 = CR(Y^2-169) 6) 27 = Y^2-169 7) 196 = Y^2 8) Y1 = 14 <=> x1 = 12 and Y2 = -14 <=> x2 = -16 And I never said, it's hard, I just see, that my "trick" was the longest to write down 8)  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post jasong jasong 4 2012-02-20 03:33 flouran Math 7 2009-12-12 18:48 davar55 Puzzles 3 2008-10-09 00:35 davar55 Puzzles 52 2007-06-26 21:41 Vijay Math 6 2005-04-14 05:19

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