Eliminating y

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  • #1
thomas49th
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For (a) I got x = [tex]\sqrt{-11}[/tex] which you can't do as you can't square root a negitive number, I can't help but feel I am wrong
 

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  • #2
cristo
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For (a) I got x = [tex]\sqrt{-11}[/tex] which you can't do as you can't square root a negitive number, I can't help but feel I am wrong

Well, for that reason I would say that Bill is wrong; not you!
 
  • #3
thomas49th
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and for (b) I began be substituting y =2x - 2 into the y²

x² + (2x-2)(2x-2) = 25

5x² -8x -21 = 0

am I doing the right thing so far. When I used the quadratic formula I got a decimal number, not a whole number and the question doesn't say anything about rounding to a degree of accuracy so I presume the answer is whole numbers :eek:
 
  • #4
arildno
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No, you may presume that the answer is to be given exactly in b)!

That is you have the two solutions:
[tex]x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4*5*(-21)}}{2*5}=\frac{8\pm\sqrt{484}}{10}=\frac{8\pm{22}}{10}[/tex]
As it happens, you get rational solutions here, otherwise, the exact solutions would involve square root symbols explicitly.
 
  • #5
cristo
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and for (b) I began be substituting y =2x - 2 into the y²

x² + (2x-2)(2x-2) = 25

5x² -8x -21 = 0

am I doing the right thing so far. When I used the quadratic formula I got a decimal number, not a whole number and the question doesn't say anything about rounding to a degree of accuracy so I presume the answer is whole numbers :eek:

What do you get when you solve this quadratic equation? The answer does not necessarily have to be a whole number, especially if it is only a short decimal. The solution to this equation is such that it can easily be written exactly.
 
  • #6
HallsofIvy
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Or, to put it another way,
5x2- 8x+ 21= (5x+ 7)(x- 3)= 0
 
  • #7
thomas49th
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i must of made an error in typing it into my calculator. For one of x's solutions I got somthing like 4.926537173 (i pressed random keys after the 3.s.f). Ill check over my work now...
 
  • #8
thomas49th
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okay, when typing it in I think I calculated the stuff inside the root wrong

anyway the answer:

x = -7 or 3
y = -16 or 4
 
  • #9
arildno
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Eeeh??
Whatever are you talking about?
 
  • #10
thomas49th
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never mind, but I've posted the (i think correct answers now above).

However I am concerned that -16² + -7² don't equal 25
 
  • #11
HallsofIvy
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arildno suggested
[tex]x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4*5*(-21)}}{2*5}=\frac{8\pm\sqrt{484}}{10}=\frac{8\pm{22 }}{10}[/tex]

and I told you that 5x2- 8x+ 21= (5x+ 7)(x- 3)= 0.

How could you possibly get "x = -7 or 3" from that?
 
  • #12
thomas49th
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(8 + 22)/10 = 3
(8-22)/10 = 1.4

woops

EDIT: Making y 4 or 4.8
 
  • #13
HallsofIvy
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Well, I would have said 7/5 but but I grew up BC (before calculators).
 
  • #14
arildno
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Actually, I would say -7/5..
 
  • #15
thomas49th
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[tex]x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4*5*(-21)}}{2*5}=\frac{8\pm\sqrt{484}}{10}=\frac{8\pm{22 }}{10}[/tex]

8+22 = 30/10 = 3
8-22 = -14/10 = -1.4

feed that into the equation y = 2(x) - 2

2(3) - 2 = 6- 2 = 4
2(-1.4) - 2 = -2.8-2 = -4.8

I would think that must be right?
 
  • #16
HallsofIvy
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Yes, that is correct. Now be sure to pair them correctly: the solutions to the pair of equations is x= 3, y= 4 and x= -1.4, y= -4.8.
 

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