anonymous
  • anonymous
taking the square root and show steps 4x^2+8x+3=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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freckles
  • freckles
sounds like you want to complete the square on one side to solve
freckles
  • 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
  • freckles
hint: \[x^2+2x+1=(x+1)^2 \]

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freckles
  • freckles
add one on both sides and try to use that last equation I posted
anonymous
  • anonymous
4x^2 + 8x + 4 = 1 (2x+2)^2 = 1 2x+2 = plus or minus 1 x = -1/2 or -3/2 ?
freckles
  • 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
  • anonymous
thank you could you help me with completing the square on the same question?
freckles
  • freckles
on the same question you already did that
freckles
  • freckles
you wrote 4x^2+8x+4 as (2x+2)^2 (2x+2)^2 is the square of (2x+2)
anonymous
  • anonymous
so taking the square and compting the square is the sam eequation?
freckles
  • 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
  • anonymous
so im suppose to show my work for completing the square and taking the root but in that i did both right?
freckles
  • 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
  • anonymous
so thats the answer for completing the square ?
freckles
  • freckles
that is the answer by using the completing the square method
freckles
  • 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
  • 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
  • anonymous
could i give that whole equation and answer all the questions in one
anonymous
  • anonymous
anonymous
  • anonymous
welll besides the factoring
freckles
  • 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
  • freckles
\[4x^2+8x+3 = 4x^2+6x+2x+3 \text{ factor this by grouping }\]
anonymous
  • anonymous
ur confusing me with all this
freckles
  • freckles
but in either method you use you should get the solutions: x=-1/2 or x=-3/2
freckles
  • freckles
with the use of quadratic formula or factoring?
anonymous
  • anonymous
both. ur talking about factoring rn or
freckles
  • freckles
what is rn?
anonymous
  • anonymous
right now
freckles
  • freckles
I was talking about both quadratic formula and factoring above:
freckles
  • 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
  • anonymous
no i have but im not sure how it works
freckles
  • freckles
can you compare: ax^2+bx+c=0 to 4x^2+8x+3=0 and identify a,b, and c?
anonymous
  • anonymous
my math program is super confusing and dosnt explain why it is
anonymous
  • anonymous
no i just need to answer these question
freckles
  • freckles
|dw:1435044171838:dw|
freckles
  • freckles
|dw:1435044201581:dw|
freckles
  • freckles
comparing your equation to that equation tells us what a,b, and c are
freckles
  • freckles
now you try pluggin into the formula I gave to find the solution we call x
anonymous
  • anonymous
-3/4,-1/2
freckles
  • freckles
the solutions shouldn't differ from the solutions you got previously
freckles
  • freckles
just the method to get the answer is a bit different
anonymous
  • anonymous
yeh i've noticed when i use mathway
anonymous
  • 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
  • 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
  • freckles
have you done the factoring and quadratic formula method yet?
freckles
  • freckles
I gave you a start on both earlier .
anonymous
  • 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
  • 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
  • anonymous
ok thank you
anonymous
  • anonymous
got to go:)

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