Find the exact solutions ofx^2 - (y- 5)^2 = 25 andy= -x^2

Do I try to isolate the y in the first equation first?

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- Jul 23rd 2007, 02:07 PMBlueStarSolving Quadratic Systems
Find the exact solutions of

*x^*2 - (*y*- 5)^2 = 25 and*y*= -*x^*2

Do I try to isolate the y in the first equation first? - Jul 23rd 2007, 02:10 PMJhevon
- Jul 23rd 2007, 02:39 PMBlueStar
Would that be like x^2 - (-x^2 - 5)^2 = 25? Or should I just drop the exponent since one is over the parenthesis?

- Jul 23rd 2007, 02:51 PMJhevon
- Jul 23rd 2007, 03:09 PMBlueStar
Is the answer 0?

x^2 - (-x^2 - 5)^2 = 25

x^2 + (x^2 + 5)^2 = 25

x^2 + x^2 + 5 = 5

x^2 + x^2 = 0

2x^2 = 0

x^2 = 0

x = 0 - Jul 23rd 2007, 03:14 PMJhevon
- Jul 23rd 2007, 03:18 PMBlueStar
So would the solutions be (0, -5)?

- Jul 23rd 2007, 03:27 PMJhevon
- Jul 23rd 2007, 03:33 PMBlueStar
http://www.mathhelpforum.com/math-he...0a18b927-1.gif

-x^4 -9x^2 = 50

??:( - Jul 23rd 2007, 03:36 PMJhevon
- Jul 23rd 2007, 03:57 PMBlueStar
http://www.mathhelpforum.com/math-he...2504fe5c-1.gifx^2 + 9x + 50 = 0

x =__-9 +- /(9)^2 - 4(1)(50)__(attempt at typing quadratic formula)

2(1)

x =__-9 +- /81 - 200__

2

But the square root number turns out to be negative... - Jul 23rd 2007, 04:08 PMJhevon
- Jul 23rd 2007, 04:14 PMBlueStar
Now I'm vaguely remembering an imaginary term,

*i*, fitting into the equation. So would it be -119*i*, or something like that? - Jul 23rd 2007, 04:27 PMJhevon
- Jul 23rd 2007, 04:35 PMBlueStar
You mean y = 9 +- 119

*i*/ -2?