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!^^