1. ## Simplify

I have the equation $x^2-x=-147$.
I played around with it but couldn't find away to get a definite answer without having to use trial and error.
Thanks!

2. Originally Posted by Truthbetold
I have the equation $x^2-x=-147$.
I played around with it but couldn't find away to get a definite answer without having to use trial and error.
Thanks!
there is good reason you had problems. this has no real solutions

if you bring everything to one side: $x^2 - x + 147 = 0$

and apply the quadratic formula: $x = \frac {1 \pm \sqrt{1 - 4(147)}}2$

obviously what is under the square root is negative, thus we have complex solutions.

3. Originally Posted by Jhevon
there is good reason you had problems. this has no real solutions

if you bring everything to one side: $x^2 - x + 147 = 0$

and apply the quadratic formula: $x = \frac {1 \pm \sqrt{1 - 4(147)}}2$

obviously what is under the square root is negative, thus we have complex solutions.

But wouldn't $x^2 - x + 147 = 0$ be in standard form so that it could be factored? $(x-?)(x+?)$

4. Originally Posted by Morzilla
But wouldn't $x^2 - x + 147 = 0$ be in standard form so that it could be factored? $(x-?)(x+?)$
Technically yes.

$x^2 - x + 147 = \left ( x - \frac{1}{2} - i~\frac{\sqrt{587}}{2} \right ) \left ( x - \frac{1}{2} + i~\frac{\sqrt{587}}{2} \right )$

As Jhevon pointed out, there are no real solutions to this problem. And the factorization is awful looking. Stick with the quadratic formula for this one; not every quadratic factors nicely.

-Dan

5. ohhhhh, so now my question is, how did we go from $
x^2 - x + 147 = 0$
this to $= \frac {1 \pm \sqrt{1 - 4(147)}}2$

what are the steps basically?

I must admit, this is getting interesting, and it's not even my problem!?!

6. Originally Posted by Morzilla
ohhhhh, so now my question is, how did we go from $
x^2 - x + 147 = 0$
this to $= \frac {1 \pm \sqrt{1 - 4(147)}}2$

what are the steps basically?

I must admit, this is getting interesting, and it's not even my problem!?!

I used the quadratic formula. it is derived by completing the square. basically it says, for any quadratic of the form $y = ax^2 + bx + c$, the roots of the quadratic (that is, the solutions to $ax^2 + bx + c = 0$) are given by:

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

7. Originally Posted by Jhevon
$y = ax^2 + bx + c$, the roots of the quadratic (that is, the solutions to $ax^2 + bx + c = 0$) are given by:

$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$
ok I remember and understand $y = ax^2 + bx + c$ and making all equal to zero, or as you mentioned, but I still can't understand the last part...
$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

I may never be able to get it

8. It's derivation is quite nice:

Cool! gracias pero, where did 4a and the $b^2$ come from!?