Math questions

Q1.
x = y = 6, so x = 6 - y. Sub into the second equation we get
(6 - y)y = 12
6y - y^2 = 12
y^2 - 6y + 12 = 0
since delta ( = b^2 - 4ac) is negative, so this equation has no real roots.
Q2.
Using the same process as Q1, we get
( 7 - y) y = 8
y^2 - 7y + 8 = 0
so y = [7 +/- sqrt(49 - 32)]/2 = (7 +/- sqrt 17)/2.
so x = 7 - y = 7 - [ 7 +/- sqrt 17]/2.

1.x+y=6
xy=12
x=6-y
(6-y)y=12
6y-y sq.=12
y sq.-6y+12=0
By formula,
y=[-(-6)(+or-)sq. root((-6)sq.-4(1)(12))]/2(1)
=[6(+or-)sq. root (-12)]/2
Since no number's sq. is negative, it has no real solution.
Therefore the equations have no real solution.

2.x+y=7
xy=8
x=7-y
(7-y)y=8
7y-y sq.=8
y sq. -7y+8=0
By formula
y=[-(-7)(+or-) sq. root((-7)sq.-4(1)(8))]/2(1)
=[7(+or-) sq. root(17)]/2
y=5.5616 or y=1.4384 (corr.to 4 d.p.)
If y =5.5616,
x=1.4384 (corr. to 4 d.p.)
if y=1.4384
x=5.5616 (corr.to 4 d.p.)

Thus the solutions are:
x=1.4384 y=5.5616
or
x=5.5616 y=1.4384

Theorem: x+y=a, xy=b if and only if x,y are roots of z^2 - az + b = 0

1. x,y are the roots of z^2 - 6z + 12 = 0
z = 3 +/- √(-3) = 3 +/- √3 i

2. x,y are the roots of z^2 - 7z + 8 = 0
z = (7 +/- √17) / 2


2009-06-01 12:47:25 補充：
1. x=3+√3 i, y=3-√3 i; or x=3-√3 i, y=3+√3 i
2. x=(7+√17)/2, y=(7-√17)/2; or x=(7-√17)/2, y=(7+√17)/2

2009-06-01 12:54:11 補充：
Remark. If you have not learnt complex number, then Q1 has no real solution.