1. ## !?!?

Solve the equation

$\displaystyle x^2+4x=3$

should I move the 3 to the left so that I can factor it or is there another way?!

Also how do I get from $\displaystyle 4n^2=-8n-1$ to $\displaystyle {\frac{-2-\sqrt{3}}{2},\frac{-2+\sqrt{3}}{2}}$

Thanks!!

2. Originally Posted by Morzilla
Solve the equation$\displaystyle x^2+4x=3$
should I move the 3 to the left so that I can factor it or is there another way?!
To move something we must be able to pick it up.
How does one pick up a number?
Do you mean add to both sides of the equation?

3. Originally Posted by Plato
To move something we must be able to pick it up.
How does one pick up a number?
Do you mean add to both sides of the equation?

hmmmm....sorry, I don't quite understand about picking up numbers.

yes, so that it'll look like this;

=$\displaystyle x^2+4x-3=3-3$

=$\displaystyle x^2+4x-3=0$

thanks

4. GOOD.
We know that we can add the same (-3) to both sides of then equation.
You are correct.

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

where:

$\displaystyle ax^2 + bx + c = 0$

6. Originally Posted by Plato
GOOD.
We know that we can add the same (-3) to both sides of then equation.
You are correct.
Ok, now that I have $\displaystyle x^2+4x-3=0$ I need to factor it out no? But I couldn't, because two numbers that when multiplied together to get -3 cannot give me +4. Since the signs are the same in $\displaystyle x^2+4x-3=0$. So then what can I do here!?

7. Originally Posted by incyt3s

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

where:

$\displaystyle ax^2 + bx + c = 0$
LOL how funny, there is another post on this formula and here it's being applied!

I will look onto the other post. I am still trying to figure out that formula!!!

Thanks

8. Originally Posted by Morzilla
Ok, now that I have $\displaystyle x^2+4x-3=0$ I need to factor it out no? But I couldn't, because two numbers that when multiplied together to get -3 cannot give me +4. Since the signs are the same in $\displaystyle x^2+4x-3=0$. So then what can I do here!?

quadratic formula. use it (or completing the square) when you can't factor something nicely

9. Originally Posted by Jhevon
quadratic formula. use it (or completing the square) when you can't factor something nicely
Until I wrote down the quadratic formula I had not realized that yes, we used this formula in class! now it's slowly coming back to me, this is where

a=x^2

b=+4x

c=-3

and then we plug it into the formula, right?!

10. Originally Posted by Morzilla
Until I wrote down the quadratic formula I had not realized that yes, we used this formula in class! now it's slowly coming back to me, this is where

a=x^2

b=+4x

c=-3

and then we plug it into the formula, right?!

not exactly. a,b, and c are the COEFFICIENTS. thus, a = 1, b = 4 and c = -3

11. Originally Posted by Morzilla
I will look onto the other post. I am still trying to figure out that formula!!!
are you familiar with the method of completing the square? if so, try to derive the formula yourself by solving the general $\displaystyle ax^2 + bx + c = 0$ by completing the square.

until then, just memorize the formula, it's not as hard as it may seem, and i know, my memory is horrible

12. Originally Posted by Jhevon
are you familiar with the method of completing the square? if so, try to derive the formula yourself by solving the general $\displaystyle ax^2 + bx + c = 0$ by completing the square.

until then, just memorize the formula, it's not as hard as it may seem, and i know, my memory is horrible
Thanks!!!