if given the eqn:
(=squared)
-x +5x +2=0

then i do the add/mult thing to find the roots, so we have to find two factors which add to give +5 and multiply to give...?
do i always multiply the +2 (c) by the coefficient of the X (a)?

therefore we have to find something which multiplies to give -2?

2. The equation is

$-x^{2}+5x+2=0,$ which I would probably write

$x^{2}-5x-2=0,$ so as to get the coefficient of the highest power of $x$ to be one. I don't think factoring is going to be the best way to solve this problem. You'd have to come up with two numbers whose product is -2, and which add up to -5. You could, of course, set up two simultaneous equations:

$uv=-2$
$u+v=-5,$ and solve. But that's adding another quadratic equation to solve, which seems a bit silly.

Have you done completing the square, or the quadratic formula yet?

3. Originally Posted by chartsy
if given the eqn:
(=squared)
-x +5x +2=0

then i do the add/mult thing to find the roots, so we have to find two factors which add to give +5 and multiply to give...?
do i always multiply the +2 (c) by the coefficient of the X (a)?

therefore we have to find something which multiplies to give -2?
first multiply through by negative 1, to make factorising easier,

$x^{2} -5x - 2 = 0$

You have to find two numbers that multiply to give you -2 and add up to give -5.

so you would have to use the quadratic equation to solve this one.

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

a = 1 , b = -5 and c = -2, put these values into the equation above and you should get two values of x,

4. lol actually i had a different forumla in my question , idon't have it by me, but can you just tell me if by factoring, the right thing to do is multiply c by the coefficient of X, even if it is negative?

5. Well, you can always multiply the entire equation by a nonzero constant without changing the solutions. You can't JUST multiply c by the coefficient of x.