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anonymous

  • one year ago

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

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  1. freckles
    • one year ago
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    sounds like you want to complete the square on one side to solve

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

  3. freckles
    • one year ago
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    hint: \[x^2+2x+1=(x+1)^2 \]

  4. freckles
    • one year ago
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    add one on both sides and try to use that last equation I posted

  5. anonymous
    • one year ago
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    4x^2 + 8x + 4 = 1 (2x+2)^2 = 1 2x+2 = plus or minus 1 x = -1/2 or -3/2 ?

  6. freckles
    • one year ago
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    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 :)

  7. anonymous
    • one year ago
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    thank you could you help me with completing the square on the same question?

  8. freckles
    • one year ago
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    on the same question you already did that

  9. freckles
    • one year ago
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    you wrote 4x^2+8x+4 as (2x+2)^2 (2x+2)^2 is the square of (2x+2)

  10. anonymous
    • one year ago
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    so taking the square and compting the square is the sam eequation?

  11. freckles
    • one year ago
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    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

  12. anonymous
    • one year ago
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    so im suppose to show my work for completing the square and taking the root but in that i did both right?

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

  14. anonymous
    • one year ago
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    so thats the answer for completing the square ?

  15. freckles
    • one year ago
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    that is the answer by using the completing the square method

  16. freckles
    • one year ago
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    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

  17. freckles
    • one year ago
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    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

  18. anonymous
    • one year ago
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    could i give that whole equation and answer all the questions in one

  19. anonymous
    • one year ago
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  20. anonymous
    • one year ago
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    welll besides the factoring

  21. freckles
    • one year ago
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    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

  22. freckles
    • one year ago
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    \[4x^2+8x+3 = 4x^2+6x+2x+3 \text{ factor this by grouping }\]

  23. anonymous
    • one year ago
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    ur confusing me with all this

  24. freckles
    • one year ago
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    but in either method you use you should get the solutions: x=-1/2 or x=-3/2

  25. freckles
    • one year ago
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    with the use of quadratic formula or factoring?

  26. anonymous
    • one year ago
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    both. ur talking about factoring rn or

  27. freckles
    • one year ago
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    what is rn?

  28. anonymous
    • one year ago
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    right now

  29. freckles
    • one year ago
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    I was talking about both quadratic formula and factoring above:

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

  31. anonymous
    • one year ago
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    no i have but im not sure how it works

  32. freckles
    • one year ago
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    can you compare: ax^2+bx+c=0 to 4x^2+8x+3=0 and identify a,b, and c?

  33. anonymous
    • one year ago
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    my math program is super confusing and dosnt explain why it is

  34. anonymous
    • one year ago
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    no i just need to answer these question

  35. freckles
    • one year ago
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    |dw:1435044171838:dw|

  36. freckles
    • one year ago
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    |dw:1435044201581:dw|

  37. freckles
    • one year ago
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    comparing your equation to that equation tells us what a,b, and c are

  38. freckles
    • one year ago
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    now you try pluggin into the formula I gave to find the solution we call x

  39. anonymous
    • one year ago
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    -3/4,-1/2

  40. freckles
    • one year ago
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    the solutions shouldn't differ from the solutions you got previously

  41. freckles
    • one year ago
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    just the method to get the answer is a bit different

  42. anonymous
    • one year ago
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    yeh i've noticed when i use mathway

  43. anonymous
    • one year ago
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    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

  44. freckles
    • one year ago
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    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 .

  45. freckles
    • one year ago
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    have you done the factoring and quadratic formula method yet?

  46. freckles
    • one year ago
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    I gave you a start on both earlier .

  47. anonymous
    • one year ago
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    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 ?

  48. freckles
    • one year ago
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    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: \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  49. anonymous
    • one year ago
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    ok thank you

  50. anonymous
    • one year ago
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    got to go:)

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