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• July 23rd 2007, 02:07 PM
BlueStar
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?
• July 23rd 2007, 02:10 PM
Jhevon
Quote:

Originally Posted by BlueStar
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?

you could, but i wouldn't. replace the y in the first equation with the -x^2 and then solve for x

once you have x, plug it back into the y = -x^2 equation to solve for y
• July 23rd 2007, 02:39 PM
BlueStar
Would that be like x^2 - (-x^2 - 5)^2 = 25? Or should I just drop the exponent since one is over the parenthesis?
• July 23rd 2007, 02:51 PM
Jhevon
Quote:

Originally Posted by BlueStar
Would that be like x^2 - (-x^2 - 5)^2 = 25? Or should I just drop the exponent since one is over the parenthesis?

no, that's correct. solve that equation for x
• July 23rd 2007, 03:09 PM
BlueStar

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
• July 23rd 2007, 03:14 PM
Jhevon
Quote:

Originally Posted by BlueStar

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

remember how to expand binomial squares: $(a \pm b)^2 = a^2 \pm 2ab + b^2$

here, you a is -x^2 and your b is -5
• July 23rd 2007, 03:18 PM
BlueStar
So would the solutions be (0, -5)?
• July 23rd 2007, 03:27 PM
Jhevon
Quote:

Originally Posted by BlueStar
So would the solutions be (0, -5)?

x = 0 is NOT a solution (if we have x = 0, that would lead to -25 = 25), and if it was, since y = -x^2, if x = 0, then y = 0, so a solution would be (0,0)

I'll start you off.

$x^2 - \left( -x^2 - 5 \right)^2 = 25$

$\Rightarrow x^2 - \left( x^4 + 10x^2 + 25 \right) = 25$

$\Rightarrow x^2 - x^4 - 10x^2 - 25 = 25$

Now continue (it seems as though the solutions are complex, as in, not real)
• July 23rd 2007, 03:33 PM
BlueStar
• July 23rd 2007, 03:36 PM
Jhevon
Quote:

Originally Posted by BlueStar

Note that we can rewrite this as: $x^4 + 9x^2 + 50 = 0$

which in turn we could rewrite as: $\left( x^2 \right)^2 + 9 \left( x^2 \right) + 50 = 0$

which is a quadratic eqaution in $x^2$, we can therefore employ the quadratic formula
• July 23rd 2007, 03:57 PM
BlueStar
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...
• July 23rd 2007, 04:08 PM
Jhevon
Quote:

Originally Posted by BlueStar
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...

yes, that's supposed to happen, as i said, this equation has complex solutions. have you done solutions like that in class?
• July 23rd 2007, 04:14 PM
BlueStar
Now I'm vaguely remembering an imaginary term, i, fitting into the equation. So would it be -119i, or something like that?
• July 23rd 2007, 04:27 PM
Jhevon
Quote:

Originally Posted by BlueStar
Now I'm vaguely remembering an imaginary term, i, fitting into the equation. So would it be -119i, or something like that?

no, it would be $x^2 = \frac {-9 \pm \sqrt {119}~i}{2}$

so now, you can solve that value for x, but you may want to plug in the expression for x^2 into y = -x^2 first
• July 23rd 2007, 04:35 PM
BlueStar
You mean y = 9 +- 119i / -2?
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