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?
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?
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?
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?