Quadratic Formula Problem

May 2010
10
0
El Cajon, CA
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
Feb 2009
3,053
1,333
West Midlands, England
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 \(\displaystyle b^2-4ac\)
 
May 2010
10
0
El Cajon, CA
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?

\(\displaystyle x = \frac{-2 \pm \sqrt{2^2-4*1*-2}}{2*1}\)
 
Feb 2010
1,036
386
Dirty South
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
 
May 2010
10
0
El Cajon, CA
\(\displaystyle x = \frac{-2 \pm \sqrt{4+8}}{2}\)

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

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

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 \(\displaystyle x = \frac{\sqrt{3}}{2}\)
 
Feb 2010
1,036
386
Dirty South
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
 
May 2010
10
0
El Cajon, CA
\(\displaystyle x = \frac{-2 + 2\sqrt{3}}{2} = \frac{2(-1 + \sqrt{3})}{2}= -1 + \sqrt{3}= \sqrt{3}-1\)

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
Sep 2008
5,237
1,625
Melbourne
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

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

If you cannot follow this please revise basic factorisation before asking additional questions.
 
May 2010
10
0
El Cajon, CA
\(\displaystyle -2+2\sqrt{3} = 2\times -1+2\times \sqrt{3}\)

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

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

If you cannot follow this please revise basic factorisation before asking additional questions.

Ok I'm still not understanding, I apologize, I'll continue to try to figure it out from my book.