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 Powering (Posted on 2013-10-05)
Given that a,b,x and y are real numbers such that

a+b=23
ax+by=79
ax2+by2=217
ax3+by3=691

Determine ax4+by4

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 possible solution | Comment 1 of 5

Let z= ax^4+by^4

I   a+b=23, b=(23-a) Substitution (b)
II  ax+by=79, a=(79-23y)/(x-y), b=(23-((79-23y)/(x-y)) Substitution (a) (b)
III  ax^2+by^2=217, ((79-23y)/(x-y))*x^2+(23-((79-23y)/(x-y)))*y^2=217, x= (217-79y)/(79-23y) Substitution (a)(b)(x)
IV  ax^3+by^3=691, the last substitution producing the heroic:
((79-23y)/(((217-79y)/(79-23y))-y))*((217-79y)/(79-23y))^3+(23-((79-23y)/(((217-79y)/(79-23y))-y)))*y^3=691
V  Simplifying spectacularly to 1250((y-3)(y+2))/(79-23y) = 0, y= {-2,3}

Substituting backwards gives x={3,-2}, a= {25,-2}, b={-2,25} when z=1993

Note: (Step V)

Let p=(79-23y), q=(217-79y)
((p)/(((q)/(p))-y))*((q)/(p))^3+(23-((p)/(((q)/(p))-y)))*y^3=691

Take LHS
q^3/(p(q-py))+23y^3+(p^2y^3)/(py-q) = q^2/p+qy+py^2+23y^3 (the term in q^3 cancels since (p^2y^3)/(py-q) = q^3/(p(py-q))+q^2/p+py^2+qy, while q^3/(p(q-py)) = -q^3/(p(py-q))

Resubstitute q (a different method could make q=(3p-20-10y))
(6241y^2-34286y+47089)/p+217y-79y^2+py^2+23y^3

Resubstitute p in the non-fractional part: 217y-79y^2+(79-23y)y^2+23y^3 = 217y
Resubstitute p throughout, put over a common denominator, and start with the numerator:
(6241y^2-34286y+47089)+(79-23y)217y
1250y^2-17143y+47089

Completing the expression:
(1250y^2-17143y+47089)/(79-23y)

Reincorporate RHS:
(1250y^2-17143y+47089)/(79-23y)-691=0

Put over a common denominator, and start with the numerator on LHS:
(1250y^2-17143y+47089)-691(79-23y)
1250y^2-17143y+47089-54589+15893y
1250y^2-(1250y)+(-7500) = 1250(y^2-y-6)
[(y^2-y-6)=(y-3)(y+2)]
Finally: 1250((y-3)(y+2))/(79-23y) [= 0]

Edited on October 5, 2013, 5:38 pm
 Posted by broll on 2013-10-05 14:24:36

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