Question

f(x)=x^2+2x-1
Find the two values of x that make f(x)=0. Round your answers to two decimal places.

Answer 1

I can't factor it, I tried the complete the square approach... I can't figure this out.

Answer 2

You can complete the square here: What will give you at least x^2 + 2x ...as a product of 2 binomials.

Answer 3

(x+1)^2, I got that. But when I continued with it. I had to find the square root of 0. which I know is not right.

Answer 4

Right and that will give you x^2 + 2x + 1, in order to make it a -1, what must you add?

Answer 5

use this
\[\Large x = \frac{b \pm \sqrt{b^{2}-4ac}}{2a}\]

Answer 6

-2.

Answer 7

Correct. Now get that to the other side of the equation and you have:
\[(x+1)^2 = 2\]
You see?

Answer 8

sorry theres - b

Answer 9

hmm... for some reason I didn't think the quadratic formula would be appropriate.... not sure why...

Answer 10

But where would the -2 come from??

Answer 11

Because you need to have a factor that will give you your original term back. -2 + 1 = -1

Answer 12

Otherwise you've completely screwed the whole equation. They are equivalent expressions.

Answer 13

The quadratic formula will also work here, of course. This is just another avenue to take.

Answer 14

I still don't totally grasp it, but it does sound familiar... is that something your supposed to do every time?? I thought when you completed the square you just added the "new number" to both sides of the equal sign.

Answer 15

You end up with the same answer finally @a.galvan2

Answer 16

Which would be easier? quadratic function? or completing the square?

Answer 17

Well thats correct, yes. (x+1)^2 = x^2 + 2x + 1, but that is not identical to the original expression. We have to turn that 1 into a -1. To do that we add a -2.

Answer 18

So we have:
\[x^2 + 2x + 1 - 2 = 0 \rightarrow (x+1)^2 -2 = 0\]