Question 1009871: In the binomial expansion of (k+ax)4 the coefficient of x2 is 54.
Given that a and k are both positive, find the value of ak.
This is my working:
(k+ax)^4 = k^4 + 4k^3ax + 6k^2ax^2 + 4kax^3
=6k^2a = 54
=k^2a = 9
ka = 3
Given also that the coefficient of x in the expansion is 48, find the value of k
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Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In the binomial expansion of (k+ax)4 the coefficient of x2 is 54.
Given that a and k are both positive, find the value of ak.
This is my working:
(k+ax)^4 = k^4 + 4k^3ax + 6k^2(ax)^2 + 4k(ax)^3 + (ax)^4
=6k^2a^2 = 54
(ka)^2 = 9
ka = 3
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Given also that the coefficient of x in the expansion is 48, find the value of k
4k^3a = 48
k^3a = 12
(ak) = 12/k^2
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Cheers,
Stan H.
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