Two maths question

1)[2(x^2)+(3/x)]^n= [3/x + 2x^2)] ^ nThe 5th term in descending powers of x = T (4+1) = (nC4) [(3/x)^(n-4)] (2x^2)^4 = (nC4) [3^(n-4) x^(4-n)] 16x^8= 16 (nC4) [3^(n-4) x^(12-n) is constant.So n = 12 ,The constant term= 16 (12C4) (3^8)= 51963120
2)a) f(3)= 4(3^2) + 3a + b= 3a + b + 36 < 0b < - 3(a + 12)
4(x^2)+ax+b = 0△ = a^2 - 4(4)b= a^2 - 16b> a^2 + 16(3(a + 12))= a^2 + 48a + 576= (a + 24)^2 > 0So f(x)=0 has two distinct real roots.

2010-08-22 11:13:02 補充：
△ > (a + 24)^2 >= 0 ,

so △ > 0

2010-08-22 11:27:00 補充：
Q1)

The 5th term in descending powers of x
= T (4+1)
= (nC4) [(2x^2) ^ (n-4)] (3/x) ^ 4
= (nC4) [2^(n-4) x^(2n-8)] 81(x^-4)

= (nC4) (81) 2^(n-4) x^(2n-12)

So n = 6 ,

Constant term

= (6C4) (81) 2^(6-4)

= 4860

2010-08-22 11:27:49 補充：
Q1)

Corrections

Q1 :

The 5th term in descending powers of x
= T (4+1)
= (nC4) [(2x^2) ^ (n-4)] (3/x) ^ 4
= (nC4) [2^(n-4) x^(2n-8)] 81(x^-4)

= (nC4) (81) 2^(n-4) x^(2n-12)

So n = 6 ,

Constant term

= (6C4) (81) 2^(6-4)

= 4860