F.4 quadratic equation

x^2 - 3x +1 = 0
Product of roots ab = 1
Sum of roots a + b = 3
a. Sum of roots (4a - 3) + (4b - 3)
= 4(a + b) - 6
= 6
Product of roots (4a - 3)(4b - 3)
= 16ab - 12a - 12b + 9
= 16 - 12(a + b) + 9
= 25 - 36
= -11
The equation is x^2 - 6x - 11 = 0
b. Sum of roots 2/a^2 + 2/b^2
= 2(a^2 + b^2) / (ab)^2
= 2(a^2 + 2ab + b^2 - 2ab)
= 2[(a + b)^2 - 2]
= 2(9 - 2)
= 14
Product of roots (2/a^2)(2/b^2)
= 4/(ab)^2
= 4
Equation is x^2 - 14x + 4 = 0
c. Sum of roots (3a^2 + 2) + (3b^2 + 2)
= 3(a^2 + b^2) + 4
= 3(a^2 + 2ab + b^2 - 2ab) + 4
= 3[(a + b)^2 - 2] + 4
= 3(9 - 2) + 4
= 25
Product of roots (3a^2 + 2)(3b^2 + 2)
= 9a^2b^2 + 6a^2 + 6b^2 + 4
= 9 + 4 + 6(a^2 + 2ab + b^2 - 2ab)
= 13 + 6[(a + b)^2 - 2]
= 13 + 42
= 55
The equation is x^2 - 25x + 55 = 0

1a) a+b=-(-3)/1=3 ab=1/1=1
From the quadratic equation with roots 4a-3 and 4b-3.
sum of roots=4a-3+(4b-3)
=4(a+b)-6
=4(3)-6
=6
products of roots=(4a-3)(4b-3)
=16ab-12b-12a+9
=16ab-12(b+a)+9
=16(1)-12(3)+9
=-11
∴The required equation is x^2-6-11=0

b)sum of roots=2/a^2+2/b^2
=2(a^2+b^2)/(ab)^2
=2[(a+b)^2-2ab]/(ab)^2
=2(9-2)/1
=14
products of roots=(2/a^2)(2/b^2)
=4/(ab)^2
=4
∴The required equation is x^2-14+4=0

c)sum of roots=3a^2 + 2+3b^2 + 2
=3(a^2+b^2)+4
=3[(a+b)^2-2ab]+4
=3(9-2)+4
=25
products of roots=(3a^2 + 2)(3b^2 + 2)
=9(ab)^2+6b^2+6a^2+4
=9+6[(a+b)^2-2ab]+4
=9+6*7+4
=55

∴The required equation is x^2-25+55=0

">>explain the change of(a^2+b^2) on b) & c)<<"
(a^2+b^2)
=a^2+2ab+b^2)-2ab
=(a+b)^2-2ab
[最難的地方只是如何轉變(a^2+b^2)]


我都是初學者~那點有錯請告訴ME......

2009-09-25 21:10:51 補充：
ANS好似差了個X
~!自己+返吧!^^