Question

Solve using any method.
2x^2 +x-5=0

Answer 1

Thats supposed to =0 not +0

Answer 2

edit your question by the little "edit" button at top

Answer 3

But anyways,

Group your x terms together with a parenthesis around them\[(2x^2+x)+5=0\]Now from the parenthesis, pull out that leading coefficient, \(2\).\[2\left(x^2+\frac{x}{2}\right) +5\]Remember, if you pull out the 2, you have to keep the equation constant by pulling it out from every variable inside the parenthesis.

Answer 4

Are you with me so far?

Answer 5

so far!

Answer 6

wait are you meaning to put +5?

Answer 7

Good :) it gets a little tricky now.

you want to "complete the square" inside the parenthesis, and you have to remember that whatever you do INSIDE the parenthesis, you must do OUTSIDE the parenthesis to keep your equation constant.

Answer 8

Sorry, meant to put \(-5\).

Answer 9

okay so how do i "complete the square"?

Answer 10

Let's rewrite that as \[2\left(x^2+\frac{x}{2}\right)-5\]

Answer 11

So now "completing the square" means pulling the equation we have inside our parenthesis into quadratic form, \(ax^2+bx+c=0\) So far, we only have \(ax^2+bx\) inside our parenthesis. We need to find our \(c\) value.

Answer 12

Still with me? :)

Answer 13

i think so..

Answer 14

\[2\left( \color{red}{x^2+\frac{x}{2}}\right)-5\]\[~~~~~~~~\downarrow\]\[~~~~~\color{red}{ax^2+bx}+c\]

Answer 15

Does that kind of clarify it a little bit more?

Answer 16

yes! thanks!

Answer 17

WOOO!!!

Answer 18

i learn better seeing!(:

Answer 19

\[b=\frac{1}{2} \implies c=\left(\frac{\dfrac{1}{2}}{2}\right)^2=\left(\frac{1}{4}\right)^2 =\frac{1}{16}\]

Answer 20

Sorry, deleted a little bit of the previous post, \[c=\left(\frac{b}{2}\right)^2\]

Answer 21

did you get the 2 from the front of the original equation after the parenthesis were added?

Answer 22

Nope, follow the post above yours :)

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
\[c=\left(\frac{b}{2}\right)^2\]
\(\color{blue}{\text{End of Quote}}\)

Answer 23

ohhh!

Answer 24

Making sense?

Answer 25

so far!

Answer 26

GREAT! (Trust me, a lot of people have trouble with this method)

Answer 27

Now we go back to our original equation, and add in that c value :)\[2\left(x^2+\frac{x}{2} +\color{blue}{\frac{1}{16}}\color{red}{-\frac{1}{16}}\right)-5\color{red}{-\frac{1}{16}}=0\]

Answer 28

To stay consistent with our original equation, like I mentioned previously, we must subtract the same value both inside and outside the parenthesis.

Answer 29

answering my question before i ever ask it!(:

Answer 30

OH NO!! MISTAKE!

Answer 31

This is something you will have to catch when writing it out,. See how there is a \(2\) factored OUTSIDE the parenthesis?

Answer 32

yeah?

Answer 33

In order to "properly" pull out the \(-\dfrac{1}{16}\), we MUST multiply it to the 2.

Just remember, everything that stays inside the parenthesis stays the same, if we pull out anything, then we must take into consideration tnhe number on the outside.

Answer 34

Therefore, it would become this: \[2\left(x^2+\frac{x}{2} +\color{blue}{\frac{1}{16}}\color{red}{-\frac{1}{16}}\right)-5\color{red}{-\frac{1}{8}}=0\] and DEFINITELY NOT \[2\left(x^2+\frac{x}{2} +\color{blue}{\frac{1}{16}}\color{red}{-\frac{1}{16}}\right)-5\color{red}{-\frac{1}{16}}=0\]See the difference?

Answer 35

okay!

Answer 36

Now we can simplify what is inside the parenthesis to \[2\left(x^2+\frac{x}{2}+\frac{1}{16}\right) - \frac{81}{16}=0\]\[2\left(x+\frac{1}{4}\right)^2-\frac{81}{16}=0\]

Answer 37

And here is a little trick. Remember: \[b=\frac{1}{2} \implies c=\left(\frac{\dfrac{1}{2}}{2}\right)^2=\left(\frac{1}{4}\right)^2 =\frac{1}{16}\] That part?

Answer 38

Tell me if I am going too fast btw.

Answer 39

yeah?

Answer 40

No, perfect speed!

Answer 41

We can simplify

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
Now we can simplify what is inside the parenthesis to \[2\left(\color{red}{x^2+\frac{x}{2}+\frac{1}{16}}\right) - \frac{81}{16}=0\]\[2\left(\color{red}{x+\frac{1}{4}}\right)^2-\frac{81}{16}=0\]
\(\color{blue}{\text{End of Quote}}\)

The highlighted portion of that formula by analyzing our \(c\) value :D \[b=\frac{1}{2} \implies c=\left(\frac{\dfrac{1}{2}}{2}\right)^2=\left(\color{red}{\frac{1}{4}}\right)^2 =\frac{1}{16}\]

That highlighted red portion of our c value will always be what our quadratic simplifies to

Answer 42

It's great because you don't have to think about it!

Answer 43

oh cool!

Answer 44

Is it making sense? :)

Answer 45

slowly but surely

Answer 46

What part are you stuck on? Let's move on from there.

Answer 47

Not stuck just wondering whats next!

Answer 48

Ohh! Ok.

Answer 49

then are we going to take away the ^2?

Answer 50

And there's that typo again -.-

Answer 51

It's funny because I'm explaining this to you and i'm making my own mistakes as I go along with this explanation.

Answer 52

It's okay! Still helping!!

Answer 53

\[2\left(x^2+\frac{x}{2} +\color{blue}{\frac{1}{16}}\color{red}{-\frac{1}{16}}\right)-5\color{red}{-\frac{1}{8}}=0\]\[2\left(x^2+\frac{x}{2}+\frac{1}{16}\right)-\frac{41}{8}=0\]\[2\left(x+\frac{1}{4}\right)^2 = \frac{41}{8}\]The reason I was getting \(-\dfrac{81}{16}\) is because I was doing what I told you not to do, and that is pulling out the \(-\dfrac{1}{16}\) from the parenthesis without multiplying it by 2.

Answer 54

Now we divide both sides of our equation by \(2\) to isolate the \(\left(x+\dfrac{1}{4}\right)^2\)

Answer 55

\[\left(x+\frac{1}{4}\right)^2=\frac{\dfrac{41}{8}}{2}\]\[\left(x+\frac{1}{4}\right)^2 =\frac{41}{16}\] Making sense?

Answer 56

yes!

Answer 57

yes!

Answer 58

Just let me know when you're ready to move on.

Answer 59

Alright :)

Answer 60

im ready (:

Answer 61

Take the square root of both sides, you end up with: \[x+\frac{1}{4} = \pm \frac{\sqrt{41}}{4}\]\[x= -\frac{1}{4}\pm \frac{\sqrt{41}}{4} = -\frac{1\pm \sqrt{41}}{4}\]

Answer 62

And if you'd like to check your answer: http://www.wolframalpha.com/input/?i=2x%5E2%2Bx-5%3D0

Answer 63

So would that be the final answer?

Answer 64

Yep :)

Answer 65

wow, thanks a lot!!

Answer 66

The alternative method is using your equation: \(2x^2+x-5=0\) into the quadratic formula: \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] where \(a = 2~,~ b = 1~,~ c = -5\)

You would also end up with \(x= \dfrac{1}{4}(-1\pm \sqrt{41})\) :)