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Solve for X

 
 
Red888
 
Reply Sun 12 Feb, 2006 09:52 am
7.1) I can get. It's simply x = 5 and y = -3

7.2) isn't as easy. y = -3/2 but check x, you can't take the square root of a negative number so what would x be? What ever it is you square it so it can't work out. Whatever is in the bracket obviously must equal 0. But now if you multipy 4 by 0 you get 0. Thats right, but then x could be any real number under the sun, because the second bracket is 0?

Why is it for 4 marks? My head is spinning.

Anyway here it is:

The Problem
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Type: Discussion • Score: 1 • Views: 524 • Replies: 8
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raprap
 
  1  
Reply Sun 12 Feb, 2006 11:10 am
1) (x-5)^2+(y+3)^2=0
if x and y are real numbers then
(x-5)^2=0 and (y+3)^2=0
why? nuttin plus nuttin is nuttin
and x=5 and y=-3 so your right here
on a real coordinate space the coordinate pair solution (x,y) is (5,-3)

2) (x^2+4)(2y+3)=0
of x and y are real then if y=-3/2
the (x^2+4)(0)=0 for any real x
on a real coordinate space the solution is a horizontal line where y=-3/2


There are other solutions if x is a complex number. A complex number includes a real and an imaginary part. such that x=a+bi, in this case a solution for x is a=o and b-2 so x=2i and since i^2=-1 then x^2=-4 and -4+4=0 and y can be any complex number.

Note with complex numbers if b=0 then the number is real

Rap
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raprap
 
  1  
Reply Sun 12 Feb, 2006 11:15 am
My grading rubric
7.1 2 points--there are two real solutions. one point for each solution
7.2 4 points--one point for recognizing that if AB=0 then A=0 or B=0--second point for the real solution for y for any real x--additional points for recognizing the complex nature of solutions for x (1 for seeing that x^2=-4 has no real solution and the other for recognizing imaginary/complex solution for x)

Rap
0 Replies
 
Lash
 
  1  
Reply Sun 12 Feb, 2006 11:46 am
We don't allow fractions here.
0 Replies
 
fresco
 
  1  
Reply Sun 12 Feb, 2006 12:26 pm
bm
0 Replies
 
fresco
 
  1  
Reply Sun 12 Feb, 2006 12:40 pm
Red888,

Can you check if there is a misprint in 7.1 please ?

These look to me like a pair of simultaneous quadratics, but substituting y=-1.5 in the first equation gives complex roots.
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Red888
 
  1  
Reply Sun 12 Feb, 2006 10:00 pm
No, I don't think it's a misprint. There are two simultaneous quadratics, question 8 and 9. This is a mixed worsheet so they don't usually have more than 2 of the same type of question. Ok thanks alot for your help all.
0 Replies
 
fresco
 
  1  
Reply Mon 13 Feb, 2006 01:20 am
Red888,

...then it is hard to see the "teaching principle" behind such examples....the first is a degenerate circle centre (5,-3) and the second, as our friends have said involves complex numbers.

BTW heres an interesting link which allows you to play with equations and watch them on a graph.

http://descartes.cnice.mecd.es/ingles/Bach_CNST_1/Simultaneous_equations_inequations/Sistemas_ecuaciones.htm

Try inserting the first equation in the last example and then alter the "0" answer to 1, 2, 3 etc....
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Red888
 
  1  
Reply Mon 13 Feb, 2006 08:23 am
They were probably trying to make us "think". Well I appreciate the help. Thanks for the link too.
0 Replies
 
 

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