Solving the Impossible: Squaring a Negative Number

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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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thomas49th said:
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!
 
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:
 
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.
 
thomas49th said:
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.
 
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...
 
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
 
never mind, but I've posted the (i think correct answers now above).

However I am concerned that -16² + -7² don't equal 25
 
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?
 
(8 + 22)/10 = 3
(8-22)/10 = 1.4

woops

EDIT: Making y 4 or 4.8
 
[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?