#### 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)

#### e^(i*pi)

MHF Hall of Honor
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 $$\displaystyle ax^2+bx+c = 0$$ then $$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

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

#### fetuslasvegas

For $$\displaystyle ax^2+bx+c = 0$$ then $$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

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

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

#### harish21

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?

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

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

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

so,

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

Both values of x satisfy your equation

#### fetuslasvegas

$$\displaystyle x = \frac{-2 \pm \sqrt{4+8}}{2}$$

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

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

so,

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

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 $$\displaystyle x = \frac{\sqrt{3}}{2}$$

#### harish21

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

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

same for the second part

#### fetuslasvegas

$$\displaystyle x = \frac{-2 + 2\sqrt{3}}{2} = \frac{2(-1 + \sqrt{3})}{2}= -1 + \sqrt{3}= \sqrt{3}-1$$

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?

#### pickslides

MHF Helper
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?
$$\displaystyle -2+2\sqrt{3} = 2\times -1+2\times \sqrt{3}$$

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

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

#### fetuslasvegas

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

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

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