What numbers multiply to 1 but add to negative 4 I have math homework on writing quadratic equations. You have to write them based on the parabola given in vertex form standard form and intercept/factored form. For the intercept form one step is to find a number that multiples to a certain number but adds to a different. So far in the equation I got to $y=-\frac{1}{2}x^2-4x+2$. I know it's not far but for the life of me I can't figure out those two numbers and without those numbers I cant do the rest. And maybe if you can, can you figure out what numbers multiply to $\frac{7}{8}$ but adds to $\frac{1}{2}$?
 A: well lets see, two numbers that multiply to 1 and add to negative 4.
Call them $a$ and $b$
$a*b = 1$  ... $(1)$
$a + b = -4$  ... $(2)$
from $(2) $
$b = -a - 4$
using $(1)$
$a*-(a+4) = 1$
$a^2 +4a + 1 = 0$    ... $(3)$
Look at that, another quadratic :)
I'll share with you a trick for solving these called completing the square. 
All quadratics can be arranged into the form $1x^2 + bx + c = 0$
As we have in the case of equation $(3)$
Once you've got that choose half of $b$
in our case thats 2
consider expanding 
$(a + 2)^2 =  \color{blue}{a^2 + 4a} + 4$
from this we can see that 
$\color{blue}{a^2 + 4a}  = (a+2)^2 - 4$
putting that back into equation $(3)$
$\color{blue}{a^2 +4a} + 1 = 0$
$\color{blue}{(a+2)^2 - 4} + 1 = 0$
$(a+2)^2 = 3$
$a = -2  \pm \sqrt{3}$
A: You can set up the equations and solve but $x= -2 - \sqrt3$ and $y=\sqrt3 - 2$ works, in regard to your second question about $7/8$ and $1/2$, I can only seem to find complex values and I am not sure if that's what you are looking for.
