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anonymous
 3 years ago
how do i factor y=0.05x^2x+1
anonymous
 3 years ago
how do i factor y=0.05x^2x+1

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you cannot, at least you cannot using rational numbers

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do you solve it with quadratic formula? i dont think my answer was right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y=0.05x^2x+1 \] \[\begin{array}{*{20}c} {x = \frac{{  b \pm \sqrt {b^2  4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in this case a=0.05 b=1 c=1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it was me, i would multiply everything by 20 and write \[x^220x+20=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0makes life a lot easier to deal with whole numbers instead of decimals

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know the formula but after i did it it looked wrong i got 1+(sr).8/.1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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 \[5x^2100x+100=0\] then divide by 5 and get \[x^220x+20=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you really do not want to take square roots of decimals or divide by decimals use whole numbers

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for \[x^220x+20=0\] it is probably easiest to complete the square because 20 is even i would write \[x^220x+20=0\] \[x^220x=20\] \[(x10)^2=20+100=80\] \[x10=\pm\sqrt{80}=4\sqrt{5}\] \[x=10\pm4\sqrt{5}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And its not factorable right?
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