anonymous
  • anonymous
how do i factor y=0.05x^2-x+1
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
you cannot, at least you cannot using rational numbers
anonymous
  • anonymous
how do you solve it with quadratic formula? i dont think my answer was right?
anonymous
  • anonymous
\[y=0.05x^2-x+1 \] \[\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array} \]

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anonymous
  • anonymous
in this case a=0.05 b=-1 c=1
anonymous
  • anonymous
if it was me, i would multiply everything by 20 and write \[x^2-20x+20=0\]
anonymous
  • anonymous
makes life a lot easier to deal with whole numbers instead of decimals
anonymous
  • anonymous
so I did it wrong?
anonymous
  • anonymous
i know the formula but after i did it it looked wrong i got 1+-(sr).8/.1
anonymous
  • anonymous
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 \[x^2-20x+20=0\]
anonymous
  • anonymous
you really do not want to take square roots of decimals or divide by decimals use whole numbers
anonymous
  • anonymous
alright thank you!
anonymous
  • anonymous
for \[x^2-20x+20=0\] it is probably easiest to complete the square because 20 is even i would write \[x^2-20x+20=0\] \[x^2-20x=-20\] \[(x-10)^2=-20+100=80\] \[x-10=\pm\sqrt{80}=4\sqrt{5}\] \[x=10\pm4\sqrt{5}\]
anonymous
  • anonymous
And its not factorable right?

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