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jkev1 Group Title

Use the quadratic formula to find the zeros of y=2x^2+8x+1

  • one year ago
  • one year ago

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  1. ParthKohli Group Title
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    \[x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}\]

    • one year ago
  2. jkev1 Group Title
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    thats not one of the options @ParthKohli

    • one year ago
  3. jkev1 Group Title
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    @Coolsector

    • one year ago
  4. jkev1 Group Title
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    @yahoo!

    • one year ago
  5. Coolsector Group Title
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    you should use what parthkohli wrote .. ax^2+bx+c =0 use what he wrote

    • one year ago
  6. jkev1 Group Title
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    it's not an option though @Coolsector

    • one year ago
  7. Coolsector Group Title
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    no .. look ax^2+bx+c =0 in your case a=2 b =8 c=1 now plug it into what parthkoli wrote and you will get the zeros ..

    • one year ago
  8. jkev1 Group Title
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    Look at this.

    • one year ago
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  9. jkev1 Group Title
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    @Coolsector

    • one year ago
  10. Coolsector Group Title
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    if you plug it into what partkholi wrote you get : \[\frac{ -8 \pm \sqrt{64-8} }{ 4 }\]

    • one year ago
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