Thread: Solving 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?

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

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

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

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

Originally Posted by BlueStar

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

remember how to expand binomial squares:

here, you a is -x^2 and your b is -5

So would the solutions be (0, -5)?

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.







Now continue (it seems as though the solutions are complex, as in, not real)

-x^4 -9x^2 = 50


??

Originally Posted by BlueStar

-x^4 -9x^2 = 50


??

Note that we can rewrite this as:

which in turn we could rewrite as:

which is a quadratic eqaution in , we can therefore employ the quadratic formula

x^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...

Originally Posted by BlueStar

x^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?

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

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

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

You mean y = 9 +- 119i / -2?