taking the square root and show steps
4x^2+8x+3=0

- anonymous

taking the square root and show steps
4x^2+8x+3=0

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- freckles

sounds like you want to complete the square on one side
to solve

- freckles

\[4x^2+8x=-3 \\ \text{ divide both sides by 4 } \\ x^2+2x=-\frac{3}{4}\]
do you recall how to factor x^2+2x+1?

- freckles

hint:
\[x^2+2x+1=(x+1)^2 \]

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## More answers

- freckles

add one on both sides and try to use that last equation I posted

- anonymous

4x^2 + 8x + 4 = 1
(2x+2)^2 = 1
2x+2 = plus or minus 1
x = -1/2 or -3/2
?

- freckles

seems great!
\[x^2+2x=\frac{-3}{4} \\ x^2+2x+1=\frac{-3}{4}+1 \\ x^2+2x+1=\frac{-3}{4}+\frac{4}{4} \\ (x+1)^2=\frac{1}{4} \\ x+1= \pm \frac{1}{2} \\ x=- 1 \pm \frac{1}{2} =\frac{-2}{2} \pm \frac{1}{2}=\frac{-2 \pm 1}{2} \\ \text{ so } x=\frac{-2+1}{2}=\frac{-1}{2} \text{ or } x=\frac{-2-1}{2}=\frac{-3}{2}\]
your way looks cute too :)

- anonymous

thank you could you help me with completing the square on the same question?

- freckles

on the same question
you already did that

- freckles

you wrote 4x^2+8x+4 as (2x+2)^2
(2x+2)^2 is the square of (2x+2)

- anonymous

so taking the square and compting the square is the sam eequation?

- freckles

I think I'm not sure what you are asking?
you solved the equation above by completing the square..
then you took square root of both sides

- anonymous

so im suppose to show my work for completing the square and taking the root but in that i did both right?

- freckles

\[(2x+2)^2 =1 \text{ this is what you got after completing the square } \\ \text{ then you took square root of both sides } \\ \sqrt{(2x+2)^2}=\sqrt{1} \\ \sqrt{(2x+2)^2}=1 \ \text{ but this means } 2x+2=\pm 1 \]

- anonymous

so thats the answer for completing the square ?

- freckles

that is the answer by using the completing the square method

- freckles

that is the answer you would also get by using any method there is to solve a quadratic such as the quadratic formula or factoring

- freckles

though it would be a bit tricky to factor the expression on the left hand side since it wouldn't be doable over integers but that doesn't mean it can't be done at all

- anonymous

could i give that whole equation and answer all the questions in one

- anonymous

##### 1 Attachment

- anonymous

welll besides the factoring

- freckles

I don't see how taking the square root and completing the square should have there on separate method
you complete the square then take square root
but anyways quadratic formula is easy you just use:
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
factoring 4x^2+8x+3 is actually not tricky (I lied ) 4(3)=6(2) and 6+2=8

- freckles

\[4x^2+8x+3 = 4x^2+6x+2x+3 \text{ factor this by grouping }\]

- anonymous

ur confusing me with all this

- freckles

but in either method you use you should get the solutions:
x=-1/2 or x=-3/2

- freckles

with the use of quadratic formula or factoring?

- anonymous

both. ur talking about factoring rn or

- freckles

what is rn?

- anonymous

right now

- freckles

I was talking about both quadratic formula and factoring above:

- freckles

\[ax^2+bx+c=0 \text{ gives } x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ is quadratic formula }\]
you never seen this formula before?

- anonymous

no i have but im not sure how it works

- freckles

can you compare:
ax^2+bx+c=0
to
4x^2+8x+3=0
and identify a,b, and c?

- anonymous

my math program is super confusing and dosnt explain why it is

- anonymous

no i just need to answer these question

- freckles

|dw:1435044171838:dw|

- freckles

|dw:1435044201581:dw|

- freckles

comparing your equation to that equation tells us what a,b, and c are

- freckles

now you try pluggin into the formula I gave to find the solution we call x

- anonymous

-3/4,-1/2

- freckles

the solutions shouldn't differ from the solutions you got previously

- freckles

just the method to get the answer is a bit different

- anonymous

yeh i've noticed when i use mathway

- anonymous

ok so
4x^2 + 8x + 4 = 1 (2x+2)^2 = 1 2x+2 = plus or minus 1 x = -1/2 or -3/2
is staking the square root
and completing the square is
(2x+2)^2
\[\sqrt{(2x+2})^2=\sqrt{1}\]
one right
sorry this isnt making ant sense

- freckles

In the completing the square method which you are already done, yes you do take the square root of both sides to solve once you got it in the form (mx+n)^2=d .

- freckles

have you done the factoring and quadratic formula method yet?

- freckles

I gave you a start on both earlier .

- anonymous

ok is this correct for the quadratic
4x^2- 6x-2x +3 = 0.
2x(x-3) - 1(2x -3) = 0
(2x -1) (2x -3) = 0
2x -1 = 0 2x - 3 = 0
2x = 1 2x = 3
x = 1/2 x = 3/2
?

- freckles

6x-2x isn't 8x
I think you meant to replace 8x with 6x+2x
4x^2+6x+2x+3=0
now try to factor the left hand side
and no you didn't use the quadratic formula
the quadratic formula is as I gave it above which was:
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

- anonymous

ok thank you

- anonymous

got to go:)

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