BMAD Posted May 7, 2013 Report Share Posted May 7, 2013 The sum of the reciprocals of two real numbers is −1, and the sum of their cubes is 4. What are the numbers? Quote Link to comment Share on other sites More sharing options...
0 witzar Posted May 7, 2013 Report Share Posted May 7, 2013 I would go with binomial formula: (x+y)3 = x3 + 3x2y + 3xy2 + y3 = (x3 + y3) + 3xy(x+y) We know that x3 + y3 = 4 and 1/x + 1/y = -1 or x + y = -xy. Let's introduce new variable: z = x + y = -xy. We can now substitute binomial formula with: z3 = 4 - 3z2, which is equivalent to (z-1)(z+2)2 = 0. Last thing to do is to solve system of equations: x + y = z xy = -z for z = 1 and for z = -2. For z=1 we get "golden ratio" solutions: (1 ± sqrt(5))/2, and z=-2 case has no real solutions. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted May 7, 2013 Report Share Posted May 7, 2013 Our old friend the golden ratio. Let: m = (x-y)/2 and n = (x+y)/2 1/m + 1/n = [2/(x-y) + 2/(x+y)] / (x2-y2) = -1 This simplifies toy2 = x2 + 4x Similarly, m3 + n3 = 4 expands and simplifies tox(x2 + 3y2) = 16 Substitute y2 from previous result to getx2(x+3) = 4 of which x=1 is clearly a root. Then from the above first result, y2=5 givesm = (1 - sqrt(5))/2 = -0.618033989 n = (1 + sqrt(5))/2 = 1.618033989 = phi = 1-m now it happens that n = -1/m so that the obvious m+n = 1 leads directly to1/m + 1/n = -1m3 = -0.2360679775n3. = 4.2360679775 whencem3 + n3 = 4 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted May 7, 2013 Report Share Posted May 7, 2013 I would go with binomial formula: (x+y)3 = x3 + 3x2y + 3xy2 + y3 = (x3 + y3) + 3xy(x+y) We know that x3 + y3 = 4and 1/x + 1/y = -1or x + y = -xy. Let's introduce new variable: z = x + y = -xy.We can now substitute binomial formula with: z3 = 4 - 3z2,which is equivalent to (z-1)(z+2)2 = 0. Last thing to do is to solve system of equations: x + y = z xy = -zfor z = 1 and for z = -2. For z=1 we get "golden ratio" solutions: (1 ± sqrt(5))/2,and z=-2 case has no real solutions. Very nice! Quote Link to comment Share on other sites More sharing options...
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BMAD
The sum of the reciprocals of two real numbers is −1, and the sum of their cubes is 4. What are the numbers?
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