binomial theorem.. in the expansion of... the coefficients bla bla bla (1 Viewer)

[studentcheese]

In the expansion of (x+a)^3 (x+b)^5, the coefficients of x^7 and x^6 are 9 and 12 respectively. Find the values of a and b.

Thanks

 

[tom.evans.15]

I just tried that question, had difficulty...

I think I compared coeffs wrong...

 

[scardizzle]

i cant be bothered writing the whole expansion out but you pretty much expand both binomials and see what multiplies together to give you x^6 and x^7

This is what i got(although my workings a bit messy so i might have missed something):

for x^7 coefficients are: 3C0 x 5C1 b + 3C1 x 5C0a

= 5b + 3a = 9

for x^6 coefficients are: 3C0 x 5C2 x b^2 + 3C1 x 5C1 x ab + 3C2 x 5C0 a^2

= 10b^2 +15ab + 3a^2 = 12

solve simultaneously

(on a side note good to see you in the forums tom)

 

[scardizzle]

by letting b be the subject of 5b + 3a = 9 ---1

you sub it into 10b^2 + 15ab + 3a^2 = 12

then you get a farely messy quadratic

the final simplified version is 4b^2 - 9b - 9

therefore b = 3 or -0.75

subbing into 1: you get a = -2 or 4.25