# One more equation

• Jul 10th 2007, 02:33 PM
earachefl
One more equation
I'm not sure how to begin attacking this one.

Given the equation $\displaystyle 4x^2 - 4xy + 1 - y^2 = 0$

use the Quadratic Formula to solve for
(a) x in terms of y (b) y in terms of x.

Can I get a hint?
• Jul 10th 2007, 03:08 PM
ThePerfectHacker
Quote:

Originally Posted by earachefl
I'm not sure how to begin attacking this one.

Given the equation $\displaystyle 4x^2 - 4xy + 1 - y^2 = 0$

use the Quadratic Formula to solve for
(a) x in terms of y (b) y in terms of x.

Can I get a hint?

Hint: $\displaystyle 4x^2 +x(-4y)+(1-y^2)=0$. So what is $\displaystyle a,b,c=?$
• Jul 10th 2007, 03:27 PM
earachefl
Hmm...

I get that I need to somehow isolate the variables on one side of the equation. I just can't seem to figure out how. It seems like every way I attack it, x and y stay mixed up together.

Are you saying to take
$\displaystyle (-4y)$ and $\displaystyle (1-y^2)$and group them together somehow?
• Jul 10th 2007, 03:34 PM
Jonboy
He's saying identify a b and c.

Remember a quadratic is in the form $\displaystyle (a)x^2\,+\,(b)x\,+\,(c)$
• Jul 10th 2007, 03:44 PM
earachefl
That was my first guess as to what he meant, and I did plug those numbers (-1, -4, 1) into the Quadratic Formula, and got the answer $\displaystyle -2 \pm \sqrt 5$. However, this is nowhere near close to the book's answers of

$\displaystyle y = -2x \pm \sqrt{8x^2 +1}$

or

$\displaystyle x = \frac {y \pm \sqrt{2y^2 - 1}}{2}$

so I'm no closer in understanding.

Am I supposed to plug the output of the Quadratic Formula in wherever y appears?
• Jul 10th 2007, 10:55 PM
Jhevon
Quote:

Originally Posted by earachefl
That was my first guess as to what he meant, and I did plug those numbers (-1, -4, 1) into the Quadratic Formula, and got the answer $\displaystyle -2 \pm \sqrt 5$. However, this is nowhere near close to the book's answers of

$\displaystyle y = -2x \pm \sqrt{8x^2 +1}$

or

$\displaystyle x = \frac {y \pm \sqrt{2y^2 - 1}}{2}$

so I'm no closer in understanding.

Am I supposed to plug the output of the Quadratic Formula in wherever y appears?

where did -1, -4, 1 come from?

let's try this hint thing one more time. i will give you a blatant one.

For (a):

$\displaystyle 4x^2 - 4xy + 1 - y^2 = 0$

$\displaystyle \Rightarrow 4x^2 + (-4y)x + \left( 1 - y^2 \right) = 0$

Here, $\displaystyle a = 4 \mbox { , } b = -4y \mbox { , } c = 1 - y^2$

For (b):

$\displaystyle -y^2 + (-4x)y + 4x^2 + 1 = 0$

Here, $\displaystyle a = -1 \mbox { , } b = -4x \mbox { , } c = 4x^2 + 1$

Can you continue now?
• Jul 11th 2007, 06:39 AM
earachefl
Thanks to all for the hints. My textbook hasn't given any example of this kind of problem, but just dumped the question in our laps regardless.
• Jul 11th 2007, 08:22 PM
Jhevon
Quote:

Originally Posted by earachefl
Thanks to all for the hints. My textbook hasn't given any example of this kind of problem, but just dumped the question in our laps regardless.

did you get the correct solutions?