Can somebody help break this problem down for me using the quadratic formula? Thanks!

x^2+2x-2=0

I know the answer is x=-1 plusminus square root 3 (don't know how to write that on here)

2. Originally Posted by fetuslasvegas
Can somebody help break this problem down for me using the quadratic formula? Thanks!

x^2+2x-2=0

I know the answer is x=-1 plusminus square root 3 (don't know how to write that on here)
For $ax^2+bx+c = 0$ then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

If you compare your equation to $ax^2+bx+c=0$ then you'll see that $a=1$, $b=2$ and $c=-2$. The place most people split up is a sign error when calculating $b^2-4ac$

3. Originally Posted by e^(i*pi)
For $ax^2+bx+c = 0$ then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

If you compare your equation to $ax^2+bx+c=0$ then you'll see that $a=1$, $b=2$ and $c=-2$. The place most people split up is a sign error when calculating $b^2-4ac$
So I know this is sort of what it would look like when I go through the process, Where do I go from there to get the answer?

$x = \frac{-2 \pm \sqrt{2^2-4*1*-2}}{2*1}$

4. Originally Posted by fetuslasvegas
So I know this is sort of what it would look like when I go through the process, Where do I go from there to get the answer?

$x = \frac{-2 \pm \sqrt{2^2-4*1*-2}}{2*1}$
$x = \frac{-2 \pm \sqrt{4+8}}{2}$

$x = \frac{-2 \pm \sqrt{12}}{2}$

$x = \frac{-2 \pm 2\sqrt{3}}{2}$

so,

$x = \frac{-2 + 2\sqrt{3}}{2} = {\sqrt{3} -1}$
OR
$x = \frac{-2 - 2\sqrt{3}}{2}= {-\sqrt{3} -1}$

Both values of x satisfy your equation

5. Originally Posted by harish21
$x = \frac{-2 \pm \sqrt{4+8}}{2}$

$x = \frac{-2 \pm \sqrt{12}}{2}$

$x = \frac{-2 \pm 2\sqrt{3}}{2}$

so,

$x = \frac{-2 + 2\sqrt{3}}{2} = {\sqrt{3} -1}$
OR
$x = \frac{-2 - 2\sqrt{3}}{2}= {-\sqrt{3} -1}$

Both values of x satisfy your equation

Ok I get mostly everything except where the -1 comes in?

The way I see the answer is $x = \frac{\sqrt{3}}{2}$

6. Originally Posted by fetuslasvegas
Ok I get mostly everything except where the -1 comes in?

The way I see the answer is $x = \frac{\sqrt{3}}{2}$
$x = \frac{-2 + 2\sqrt{3}}{2} = \frac{2(-1 + \sqrt{3})}{2}= -1 + \sqrt{3}= \sqrt{3}-1$

same for the second part

7. Originally Posted by harish21
$x = \frac{-2 + 2\sqrt{3}}{2} = \frac{2(-1 + \sqrt{3})}{2}= -1 + \sqrt{3}= \sqrt{3}-1$

same for the second part
I see how it makes sense, I just am not getting where this -1 is coming from or what you are doing to get the -1. Is it coming from moving the negative sign from the first two which for some reason subtracts from the second 2?

8. Originally Posted by fetuslasvegas
I see how it makes sense, I just am not getting where this -1 is coming from or what you are doing to get the -1. Is it coming from moving the negative sign from the first two which for some reason subtracts from the second 2?
$-2+2\sqrt{3} = 2\times -1+2\times \sqrt{3}$

Now 2 is common to both terms so it can be factored out giving

$2\times -1+2\times \sqrt{3} = 2(-1+\sqrt{3})$

$-2+2\sqrt{3} = 2\times -1+2\times \sqrt{3}$
$2\times -1+2\times \sqrt{3} = 2(-1+\sqrt{3})$