F.2 - F.3 MATH (20x2=40)

1. For the expression a(x^2) +bx+ 25, write down two possible sets of the values of a and b such that the expression is a perfect square.

discriminant= b^2 -4a(25)=0
b^2 -100a=0


2. {(6x)/[3(x^2) 3x]}+{(x-1)/[2(x 1)^2]} - {(x)/[6(x^2)-6]}

={(6x)/[3x(x+1)]}+{(x-1)/[2(x+1)^2]} - {(x)/[6x(x-1)]}
={2/[(x+1)]}+{(x-1)/[2(x 1)^2]} - {1/[6(x-1)]}
=[12(x+1)(x-1)+3(x-1)^2- (x+1)^2]/[6(x-1)(x+1)^2]}
=[12(x^2-1)+3(x^2-2x+1)- (x^2-2x+1)]/[6(x-1)(x+1)^2]}
=[14x^2- 4x-10]/[6(x-1)(x+1)^2]
=[7x^2- 2x-5]/[3(x-1)(x+1)^2]
=[(7x-5)(x+1)]/[3(x-1)(x+1)^2]
=(7x-5)/[3(x-1)(x+1)]

3. For the simultaneous equations x+y=3 and 3x+3y=9, Rachel suggests that it has infinite number of solutions. Do you agree with her? Explain your answer briefly.

Yes, the reason is that the equations represent the same straight line. This means that the lines collapse as one. So all the points on the line are the solutions

4. Set up a pair of simultaneous linear equations in two unknowns that have no solutions.
Parallel lines have no intersections so the simultaneous linear equations set by those lines will have no solutions.

x+y=3 and x+y=7

2008-08-02 05:22:59 補充：
真正做應該用 completing square,
假設有一expression
a(x^2) +bx+ c
=a[(x^2) +(b/a)x+ c/a]
=a｛(x^2) +(b/a)x+ [√(c/a)]^2｝
=a｛(x^2) +[√(c/a)]x+ [√(c/a)]^2 +(b/a)x -2[√(c/a)]x｝
=a｛(x^2) +[√(c/a)]x+ [√(c/a)]^2｝ +｛(b/a)x -2[√(c/a)]x｝
=a｛(x+ [√(c/a)]｝^2 +｛(b/a) -2[√(c/a)]｝x
｛(b/a) -2[√(c/a)]｝Must =0
字數有限!

2008-08-02 05:27:56 補充：
when
(b/a)-2(√c/a) =0
(b/a)=2(√c/a)
(b/a)^2=4(c/a)
b^2=4ac
b^2-4ac =0
this is called discriminant

2008-08-03 04:09:47 補充：
之前有少少錯誤, 以下才是正確
真正做應該用 completing square,
假設有一expression
a(x^2) +bx+ c
=a[(x^2) +(b/a)x+ c/a]
=a[(x^2) +2(1/2)(b/a)x+ c/a] < ---由這兒開始
=a｛(x^2) +2(1/2)(b/a)x +[(1/2)(b/a)]^2 -[(1/2)(b/a)]^2 +c/a｝
=a｛ [x+(1/2)(b/a)]^2 -[(1/2)(b/a)]^2 +c/a｝< ---後面那堆就是 discriminant
[(1/2)(b/a)]^2 +c/a

2008-08-03 04:19:04 補充：
-[(1/2)(b/a)]^2 +c/a
=-(b^2/4a^2) +c/a
=-(b^2/4a^2) +(4ac/4a^2)
=(-b^2+4ac/4a^2)

所以, 只要 -b^2+4ac=0
a(x^2) +bx+ c 就有條件成為perfect square