What numbers multiply to 1 but add to negative 4

Question

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 $\frac{1}{2}$?

Answer 1

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.

Answer 2

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 $(3)$

$\color{blue}{a^2 +4a} + 1 = 0$

$\color{blue}{(a+2)^2 - 4} + 1 = 0$

$(a+2)^2 = 3$

$a = -2 \pm \sqrt{3}$