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anonymous

  • one year ago

hi

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  1. anonymous
    • one year ago
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    what is your question about this quadratic equation

  2. anonymous
    • one year ago
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    x(x-4)=0 x=0 or x-4=0 x=0 or x=4

  3. Nnesha
    • one year ago
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    find the GCF what is common in both terms ?

  4. anonymous
    • one year ago
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    i need help using the quadratic formula for this problem........

  5. Nnesha
    • one year ago
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    oaky should mention that *using the qudratic formula*

  6. anonymous
    • one year ago
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    sorry lol.........

  7. Nnesha
    • one year ago
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    \[\huge\rm x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] plugin a ,b c values \[\huge\rm Ax^2+Bx+C=0\] a=leading coefficient b= middle term c= constant term

  8. anonymous
    • one year ago
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    \[\frac{ 4\sqrt{16 - 0} }{ 2}\]

  9. Nnesha
    • one year ago
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    a= ? b=? c= ?

  10. anonymous
    • one year ago
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    actually nvm listen to @Nnesha

  11. anonymous
    • one year ago
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    damn i think i just gave her the answer ... sorry about that @Nnesha

  12. Nnesha
    • one year ago
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    \[\huge\rm \color{reD}{A}x^2+\color{blue}{B}x+\color{green}{C}=0\] \[\huge\rm \color{reD}{1}x^2\color{blue}{-4}x=0\]a ,b ,c values are a= leading coefficient b=middle term c=constant term

  13. Nnesha
    • one year ago
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    so what is a , b and c in this equation 1x2 -4x = 0 ??

  14. anonymous
    • one year ago
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    this is wat i got so far,,, \[\frac{ (-4±\sqrt{-4^2-4*0*1}) }{ 2*1}\]

  15. anonymous
    • one year ago
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    i know how to substitute and all..its the solving part......... @Nnesha

  16. Nnesha
    • one year ago
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    put the parentheses (-4)^2

  17. anonymous
    • one year ago
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    i forgot that

  18. anonymous
    • one year ago
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    oops ok next.........

  19. Nnesha
    • one year ago
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    -4^2 isn't same as (-4)^2 and there is negative sign in the formula and b value is also negative \[\huge\rm \frac{\color{ReD}{-} (-4)±\sqrt{(-4)^2-(4*0*1)} }{ 2*1}\]

  20. anonymous
    • one year ago
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    \[\frac{ (−(-4)±\sqrt{16−4∗0∗1}) }{ 2 }\]

  21. Nnesha
    • one year ago
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    yep :)

  22. Nnesha
    • one year ago
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    -(-4) distribute :)

  23. anonymous
    • one year ago
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    \[\frac{ 4±\sqrt{1} }{ 2 }\]

  24. anonymous
    • one year ago
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    ???

  25. Nnesha
    • one year ago
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    hmmm

  26. anonymous
    • one year ago
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    then??...

  27. anonymous
    • one year ago
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    \[\frac{ 4\pm1 }{ 2 }\]

  28. anonymous
    • one year ago
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    \[\frac{ 4+1 }{ 2 }=\frac{ 5 }{ 2 }\]

  29. Nnesha
    • one year ago
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    how did you get one under the square root ?

  30. anonymous
    • one year ago
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    \[\frac{ 4-1 }{ 2 }=\frac{ 3 }{ 2 }\]

  31. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @plzzhelpme \[\frac{ 4±\sqrt{1} }{ 2 }\] \(\color{blue}{\text{End of Quote}}\) here how did you get one ? :)

  32. anonymous
    • one year ago
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    i subtracted 4 from 16 = 12 then 12*0*1=1

  33. anonymous
    • one year ago
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    ohhh i did that part wrong.......

  34. anonymous
    • one year ago
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    its 15??

  35. Nnesha
    • one year ago
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    ye ^^ try again \[\sqrt{16-(4 \times 0 \times 1)}\]

  36. Nnesha
    • one year ago
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    no not 15

  37. Nnesha
    • one year ago
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    when ou multiply a number by 0 what will you get ?

  38. Nnesha
    • one year ago
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    4 times 0 ? x times 0 = ?? chocolates times 0 ?

  39. anonymous
    • one year ago
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    0

  40. Nnesha
    • one year ago
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    yes so 4 times 0 times 1 = 0 \[\sqrt{16-(4 \times 0 \times 1)}\] \[\sqrt{16-0}\]

  41. anonymous
    • one year ago
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    16

  42. Nnesha
    • one year ago
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    sqrt{16} = ?

  43. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @plzzhelpme \[\frac{ 4±\sqrt{1} }{ 2 }\] \(\color{blue}{\text{End of Quote}}\) so it's \[x=\frac{ 4±\sqrt{16} }{ 2 }\] take square root of 16 and then solve!

  44. anonymous
    • one year ago
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    oh ok.....thx.....

  45. Nnesha
    • one year ago
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    got the answer ?

  46. anonymous
    • one year ago
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    ya 0 and 4

  47. Nnesha
    • one year ago
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    :=)

  48. anonymous
    • one year ago
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    @Nnesha how would u check the solution when we only have : x=4 ?

  49. Nnesha
    • one year ago
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    plug it into the equation if both sides are equal then yes 4 is a solution of that equation and if both sides are not equal then nope 4 isn't a solution of that equation

  50. anonymous
    • one year ago
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    i got this https://www.symbolab.com/solver/step-by-step/4%5E%7B2%7D-4%5Cleft(4%5Cright)%3D0/?origin=enterkey @Nnesha

  51. Nnesha
    • one year ago
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    let 's use our own pretty hands 4^32-4(4) = 0 16-16=0 both sides are equal r?

  52. Nnesha
    • one year ago
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    hurry up! gtg ;P

  53. anonymous
    • one year ago
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    why 4^32 ?? @Nnesha

  54. Nnesha
    • one year ago
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    how ??

  55. Nnesha
    • one year ago
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    ohh my bad *backspace* button isn't working only on this page so yea

  56. anonymous
    • one year ago
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    ok ya thats 0

  57. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha let 's use our own pretty hands 4^32-4(4) = 0 16-16=0 both sides are equal r? \(\color{blue}{\text{End of Quote}}\) correction 4^2 -4(4)=0

  58. Nnesha
    • one year ago
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    wiill give you 16-16=0 (16-16) =0 so 0=0 both sides are equal so yes 4 is a solution

  59. anonymous
    • one year ago
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    yayyy thx soo much........

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