Solve the equation 4x2 + 8x + 1 = 0 by completing the square.

- some.random.cool.kid

Solve the equation 4x2 + 8x + 1 = 0 by completing the square.

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- some.random.cool.kid

@Hero @dan815 @iambatman

- some.random.cool.kid

@phi @amistre64

- some.random.cool.kid

@Nnesha mind helping :D

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## More answers

- some.random.cool.kid

not to much this time...

- some.random.cool.kid

@Nnesha

- anonymous

First, make the coefficient of the x^2 term 1 by dividing both sides by 4.

- some.random.cool.kid

ok

- some.random.cool.kid

so 4 by which sides?

- anonymous

Divide all terms on both sides of the equation by 4.

- some.random.cool.kid

the x^2 and the 8x by 4?

- anonymous

Everything, yes. What will it look like?

- some.random.cool.kid

uh... well 8x would be 2x?

- anonymous

Yup. Keep going.

- some.random.cool.kid

and what about the ^2

- some.random.cool.kid

is that 4^2?

- anonymous

Divide 4x^2 by 4. What do you get?\[\frac{ 4x^2 }{ 4 } = ?\]

- some.random.cool.kid

just x^2?

- anonymous

Right. Keep going.

- some.random.cool.kid

1 divided by 4 is that 4?

- anonymous

No. It's \(\frac{1}{4}\). What about the right hand side? Gotta do that too.

- some.random.cool.kid

0

- anonymous

Good. So, dividing everything by 4 gives\[x^2 + 2x + \frac{ 1 }{ 4 } = 0\]OK so far?

- some.random.cool.kid

alright

- anonymous

Now, we don't want the 1/4 on the left hand side, so subtract 1/4 from both sides. What will it look like then?

- some.random.cool.kid

uh - 1/4?

- some.random.cool.kid

thats a negative

- anonymous

OK. What does the whole equation look like now?

- some.random.cool.kid

x^2 + 2x - 1/4?

- anonymous

Not quite. What happened to the equals sign?

- some.random.cool.kid

idk is it x^2 + 2x - 1/4 =0
or
x^2 + 2x = 1/4?

- anonymous

We better back up a step. You had\[x^2 + 2x + \frac{ 1 }{ 4 } = 0\]To get rid of the 1/4 on the left hand side, you need to subtract it. And the rules of equality say that whatever you do one side of the equation you have to do to the other side as well. So you have to subtract 1/4 from both sides. So\[x^2 + 2x + \frac{ 1 }{ 4 } - \frac{ 1 }{ 4 } = 0-\frac{ 1 }{ 4 }\]Simplify both sides. What will it look like?

- some.random.cool.kid

wait but if you canceled the other 1/4ths and your left - 1/4 then what is left i dont get it all i se is x^2 + 2x - 1/4

- anonymous

This is not an equation. What happened to the equal sign. It can't just disappear.

- some.random.cool.kid

you dont have one you used it when you crossed out the 1/4ths

- some.random.cool.kid

thats how i was tought...

- anonymous

Sorry, no. Equals signs don't disappear. What you are left with is\[x^2 + 2x = -\frac{ 1 }{ 4 }\]Do you understand?

- some.random.cool.kid

ooo ok

- anonymous

OK. So now you're ready to complete the square on the left hand side. Do you remember how to do it?

- some.random.cool.kid

some more dividing? lawl

- anonymous

Not really :) Do you know how to complete the square?

- some.random.cool.kid

mm take half the sq on both sides?

- anonymous

Something like that. You need to add a number to the left hand side that will make it a perfect square trinomial. Look at the coefficient of the 'x' term. Take half of that coefficient and then square it. What number do you get?

- some.random.cool.kid

your left with x +2? after taking away one x from each of the sides?

- anonymous

No. What is the coefficient of the 'x' term on the left hand side?

- some.random.cool.kid

sorry idk that..

- anonymous

What number is multiplying x on the left hand side?

- some.random.cool.kid

1

- anonymous

Oh my. The equation you're working with now is\[x^2 + 2x = -\frac{ 1 }{ 4 }\]The 'x' term on the left hand side is \(2x\). The coefficient is the number that is multiplying the \(x\). What number is it?

- some.random.cool.kid

omg i said idk geez...

- anonymous

|dw:1443484097406:dw|

- anonymous

In \(2x\), the \(x\) is being multiplied by \(2\). The 2 is called the coefficient of x. Do you understand?

- some.random.cool.kid

that is a variable you cant multiply that :/

- some.random.cool.kid

you have to simplify it...

- anonymous

Sure you can. 2 times x is 2x. 14 times y is 14y. And so on.

- some.random.cool.kid

x equals start fraction three plus or minus square root of six end square root over two end fraction
x equals six plus or minus two square root of six end square root
x equals start fraction negative two plus or minus square root of three end square root over two end fraction
x equals start fraction four plus or minus three square root of six end square root over eight end fraction

- anonymous

So the coefficient of the 'x' term is 2. Now take half of the coefficient - half of 2 is 1. Now square that number. What is 1 squared?

- some.random.cool.kid

1

- some.random.cool.kid

anyways those are what i have...

- some.random.cool.kid

im thinking its bewtween those...

- anonymous

Very good. So you need to add 1 to both sides of the equation\[x^2 + 2x = -\frac{ 1 }{ 4 }\]What will the equation look like?

- some.random.cool.kid

like this in the end... x equals start fraction four plus or minus three square root of six end square root over eight end fraction

- anonymous

How do you get that?

- some.random.cool.kid

out of these and based on your thing this was best choice.
x equals start fraction three plus or minus square root of six end square root over two end fraction
x equals six plus or minus two square root of six end square root
x equals start fraction negative two plus or minus square root of three end square root over two end fraction
x equals start fraction four plus or minus three square root of six end square root over eight end fraction

- anonymous

If you would like my help in learning how to complete the square, I'll help you. But, if you're just going to copy someone else's work (which hasn't been reduced to lowest terms, by the way), then I'll move on. Good luck with your work.

- some.random.cool.kid

this is my work lol

- some.random.cool.kid

these were my answers and your answer werent near what i had...

- some.random.cool.kid

@Nnesha

- anonymous

You haven't completed the work required to arrive at an answer.

- some.random.cool.kid

the way you were headed i figured i would paste my answers because it seemed like you were going in a different direction... as soon as you said i can multiply i knew it was off because in this course you do not multiply these it strictly says you dont multiply them its supposed to be simplified @ospreytriple

- some.random.cool.kid

the variables i mean...

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