Simplify

**Truthbetold** · December 13th 2007, 07:36 PM

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!

**Jhevon** · December 13th 2007, 08:54 PM

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.

**Morzilla** · December 15th 2007, 06:16 AM

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? $(x-?)(x+?)$

**topsquark** · December 15th 2007, 06:45 AM

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

**Morzilla** · December 15th 2007, 05:55 PM

ohhhhh, so now my question is, how did we go from $<br /> x^2 - x + 147 = 0$ 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!?!

**Jhevon** · December 15th 2007, 05:58 PM

Originally Posted by Morzilla

ohhhhh, so now my question is, how did we go from $<br /> x^2 - x + 147 = 0$ 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:

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

**Morzilla** · December 16th 2007, 07:16 AM

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

**DivideBy0** · December 16th 2007, 08:41 AM

It's derivation is quite nice:

Quadratic Formula Derivation

But you will need to know how to complete the square. It is a necessary prerequisite.

This is hard stuff, so take your time with it and don't get yourself down if you can't understand it after a while.

**Krizalid** · December 16th 2007, 09:26 AM

Here I attach another solution.

**Morzilla** · December 16th 2007, 09:41 AM

Originally Posted by Krizalid

Here I attach another solution.

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

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