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fluffyunicorn

  • 2 years ago

how do i factor y=0.05x^2-x+1

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  1. satellite73
    • 2 years ago
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    you cannot, at least you cannot using rational numbers

  2. fluffyunicorn
    • 2 years ago
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    how do you solve it with quadratic formula? i dont think my answer was right?

  3. Monkeyball
    • 2 years ago
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    \[y=0.05x^2-x+1 \] \[\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array} \]

  4. Monkeyball
    • 2 years ago
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    in this case a=0.05 b=-1 c=1

  5. satellite73
    • 2 years ago
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    if it was me, i would multiply everything by 20 and write \[x^2-20x+20=0\]

  6. satellite73
    • 2 years ago
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    makes life a lot easier to deal with whole numbers instead of decimals

  7. Monkeyball
    • 2 years ago
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    so I did it wrong?

  8. fluffyunicorn
    • 2 years ago
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    i know the formula but after i did it it looked wrong i got 1+-(sr).8/.1

  9. satellite73
    • 2 years ago
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    no it is just easier if it is not clear where i got the 20 from, lets clear the decimal by multiplying everything by 100 you get \[5x^2-100x+100=0\] then divide by 5 and get \[x^2-20x+20=0\]

  10. satellite73
    • 2 years ago
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    you really do not want to take square roots of decimals or divide by decimals use whole numbers

  11. fluffyunicorn
    • 2 years ago
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    alright thank you!

  12. satellite73
    • 2 years ago
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    for \[x^2-20x+20=0\] it is probably easiest to complete the square because 20 is even i would write \[x^2-20x+20=0\] \[x^2-20x=-20\] \[(x-10)^2=-20+100=80\] \[x-10=\pm\sqrt{80}=4\sqrt{5}\] \[x=10\pm4\sqrt{5}\]

  13. fluffyunicorn
    • 2 years ago
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    And its not factorable right?

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