BINOMIAL THEOREM

23) if (1+ax+bx^2)^n = 1 + 7x - 7x^2 + terms involving higher powers of x,where a and b are integers,find the values of a ,b,and n ,
(1+ax+bx^2)^n
=(1+x(a+bx))^n
=1+nx(a+bx)+nC2[x(a+bx)]^2+ terms involving higher powers of x
=1+nx(a+bx)+nC2(x^2)(a^2+2abx+x^2)+ terms involving higher powers of x
=1+anx+(bn+nC2(a^2))x^2+terms involving higher powers of x
So
an=7
bn+nC2(a^2)=-7
we know that n is a positive number and 7 is a prime number
So
n=7, a=1
That is
7b+21=-7
b=-4
24) (a) prove that (a+b)^4-4ab(a+b)^2 + 2a^2b^2 = a^4+b^4
(b) given that a is a positive number and b is a negative number satisfyying a+b=1 and a^4+b^4 =97.using the result of (a) ,show that ab= -6.
(c) hence,find the values of a and b.
(a)
LHS
= (a+b)^4-4ab(a+b)^2 + 2a^2b^2
=a^4+4a^3b+6a^2b^2+4a^3b+b^4-4ab(a^2+2ab+b^2)+2a^2b^2
=a^4+4a^3b+6a^2b^2+4a^3b+b^4-4a^3b-8a^2b^2-4ab^2^3+2a^2b^2
=a^4+b^4
=RHS
(b)
a+b=1 and a^4+b^4 =97
Using (a+b)^4-4ab(a+b)^2 + 2a^2b^2 = a^4+b^4
1-4ab+2a^2b^2 = 97
-96=4ab-2a^2b^2
-48=2ab-a^2b^2
ab(ab-2)=48
Since 48=-6*-8
we conclude that
ab=-6
(c)
ab=-6
a+b=1 (given)
a=3, b=-2 (using trial and error or substitution method)