Solve the formula for y?
x^2+y^2=r^2
I subrtracted x^2 from both sides and then divided y^2 on both sides...am I on the right track or totally off?
Change the equation to Quadratic Form and then solve using any method?
x(2x+1)/x-4=36/x-4
Solve the formula for y?
x^2+y^2=r^2
I subrtracted x^2 from both sides and then divided y^2 on both sides...am I on the right track or totally off?
Change the equation to Quadratic Form and then solve using any method?
x(2x+1)/x-4=36/x-4
1. Given $\displaystyle x^2 + y^2 = r^2$:
Subtract $\displaystyle x^2$ from both sides.
$\displaystyle y^2 = r^2 - x^2$
Take the square root of both sides.
$\displaystyle y = \pm\sqrt{r^2 - x^2}$
2. Given $\displaystyle \frac{x(2x + 1)}{x - 4} = \frac{36}{x - 4}$:
Multiply both sides by x - 4.
$\displaystyle x(2x + 1) = 36$
Distribute x and subtract 36 from both sides.
$\displaystyle 2x^2 + x - 36 = 0$
Solve using the quadratic formula.
Here are the general rule. If your quadratic equation is in the form $\displaystyle ax^2+bx+c =0$, then for:
Completing the square - $\displaystyle a\left[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right]=0$.
Quadratic equation - $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.