Question

u^2-4u-1=0

Answer 1

What are we doing here?

Answer 2

solving quadratic equations. I just don't know if I need to factor it first

Answer 3

Oh, it won't factor.

Answer 4

So you can use the quadratic formula or you can solve by completing the square.

Answer 5

can you show me im having a hard time with this math 230 class

Answer 6

You got it! :)

Answer 7

thank you

Answer 8

Now, which way would you like me to show you? By using the quadratic formula, or completing the square?

Answer 9

how about completing the square that seems to be the hardest part

Answer 10

Sure thing!

Answer 11

To get started, we'll write out the formula and identify the pieces:

The first thing to not is that right now it is in standard form.\[u^2-4u-1=0\]
To solve by completing the square, we don't want it in standard form though. The reason we don't want it in standard form is because we're trying to create a perfect square trinomial and our current "c" term does not guarantee we will have a perfect square trinomial. It's the "bad c".

Answer 12

ok

Answer 13

And the "bad c" needs to be moved to the other side of the equation. So we'll add 1 to both sides of this equation:
\[u^2-4u=1\]

Answer 14

Now, I'm going to do a rewrite, but I want to leave room for the "good c" that we're about to create. So here's what it looks like now:
u^2-4u+___=1+___

Answer 15

ok so far so good lol

Answer 16

I left 2 blank spaces for the number we're about to add to both sides.

Answer 17

this was part of what i didnt understand where they got this number from

Answer 18

Now, to find "good c" we take our b value, chop it in half, and square it. Like this:
\[(\frac {b}{2})^2\]

Answer 19

oh yeah i kinda remember doing this before

Answer 20

\[c=(\frac{b}{2})^2\]Now, we'll plug in our 'b' value:
\[c=(\frac{(-4)}{2})^2\]
\[c=(-2)^2\]
c=4

Answer 21

So, we've discovered "good c" is 4 and that'll be the number we fill those blanks in with.

Answer 22

So here's what we have so far:
\[u^2-4u+4=1+4\]Now, we want to factor the left side and combine terms on the right side.

Answer 23

ok got it so far

Answer 24

So here's what the left side is factored, and the right side is combined:
\[(x-2)^2=5\]

Answer 25

From here, we solve using the square root method:
\[\sqrt{(x-2)^2}=\sqrt{5}\]And when we square root both sides of an equation, we need to consider the plus or minus.
\[(x-2)=\pm \sqrt{5}\]Then we add 2 to both sides, and when we write our final answer the plus/minus piece comes second. Like this:
\[x=2\pm \sqrt{5}\]

Answer 26

Those are your exact answers. If you needed approximate answers, for whatever reason, you could run those through a calculator. :)

Answer 27

Thank you so much you were a lot of help. You made it sound so easy

Answer 28

Did that make sense?

Answer 29

Yay!! :)

Answer 30

yeah a lot more than all the videos i have been watching

Answer 31

Well, I'm happy to help.

Answer 32

I'm going to get a metal, right? ;)

Answer 33

yeah how do i do that new to this site

Answer 34

Perfect! :) Thanks!!

Answer 35

i figured it out lol

Answer 36

one more question though what happens if i can factor it like\[x ^{2}+x+6=0\]

Answer 37

I factored it should i have done that