# c2 question, help pleaseWatch

Thread starter 12 years ago
#1
a) given that (2+5)^5+(2-x)^5=A+Bx^2+Cx^4
Find constants of A, B, and C
So I expanded it and found..
A=54, B=160, C=20

b) using the substitution y=x^2 and your answers to part a, solve
(2+x)^5+(2-x)^5=349

help on part b please

Thanks very much
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12 years ago
#2
Hi, I didn't check if A, B and C are correct, but if they are here's what you got to do: solve the equation 54+160y+20y^2=349 once you found the solutions to this equation, come back to y=x^2 and thus you can find x (remember to not forget some solutions i.e if y is positif then there are two different x's of opposite sign that satisfy the equation y=x^2, if y is negatif then I don't think you're supposed to know how to deal with the equation so just say that the square root of a negative number doesn't exist)
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12 years ago
#3
Surely A = 64
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12 years ago
#4
(Original post by Libertine)
Surely A = 64
Agreed.

64+160x^2+20x^4 = 349
64+160y+20y^2=349

Solve for y and then sub back into y=x^2 for x values.
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Thread starter 12 years ago
#5
A is 64 sorry my mistake and thanks everyone
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