Finding x-intercepts - how is this done? (Quadratic Formula)

Hey folks, wanted to say thanks again for the help thats been given to me.

I was doing my homework and I got a little stuck on something. Confused on how to use this'quadratic forumula'

I used it a few times with the sample questions with no problems, but with this one there seems to be a step I'm not comprehending in the sample answer.

Heres the question: Solve *x*^{2} + 2*x* = 1. Round to two decimal places.

I know I'm supposed to get the right side to equal zero, and I know how to sub in the numbers in the formula but there was one specific step in this sample thats got me stumped.

Heres the given answer:

- Letting
*a* = 1, *b* = 2, and *c* = –1, the Quadratic Formula gives me:

- http://www.purplemath.com/modules/quads/qform09.gif

I get lost in the second row. I don't understand why the 8 in the square sudden turns to a 4, and a 2. Why would I do this, and how do I know when I'm supposed to do it when working with this formula?

This is from an online course im doing, and ofcourse there is no explanation of how it was done, but they explain something as simple and moving the 1 to the left side to make the right side equal zero.

Re: Finding x-intercepts - how is this done? (Quadratic Formula)

The quadratic formula says that the two solutions to the quadratic equation $\displaystyle ax^2+ bx+ c= 0$ are given by $\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$. I can't see that attachment- just "Purplemath Graphic" so I can't tell what is wrong. Here the equation is $\displaystyle x^2- 2x- 1= 0$. a= 1, b= -2, c= -1 so that becomes $\displaystyle x= \frac{-(-2)\pm\sqrt{(-2)^2- 4(1)(-1)}}{2(-1)}$$\displaystyle = \frac{2}\pm\sqrt{4+ 4}{-2}= -\frac{2\pm\sqrt{8}}{2}$

Ah, now I see what your question is. They are using the fact that 8= 4(2) so that $\displaystyle \sqrt{8}= \sqrt{4(2)}=\sqrt{4}\sqrt{2}= 2\sqrt{2}$ because $\displaystyle \sqrt{4}= 2$. So the formula reduces to $\displaystyle -\frac{2\pm 2\sqrt{2}}{2}= -1\pm \sqrt{2}$.

Re: Finding x-intercepts - how is this done? (Quadratic Formula)

Re: Finding x-intercepts - how is this done? (Quadratic Formula)

Quote:

Originally Posted by

**Urpes** (Swear) thats difficult...

You get used to it. Please note that HallsofIvy has a slight typo....b = 2, not -2 (and for some reason a = -1 in the denominator, but not under the square root.). The final answer he gives is correct, though.

-Dan