Question

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

Answer 1

you cannot, at least you cannot using rational numbers

Answer 2

how do you solve it with quadratic formula? i dont think my answer was right?

Answer 3

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

Answer 4

in this case
a=0.05
b=-1
c=1

Answer 5

if it was me, i would multiply everything by 20 and write
\[x^2-20x+20=0\]

Answer 6

makes life a lot easier to deal with whole numbers instead of decimals

Answer 7

so I did it wrong?

Answer 8

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

Answer 9

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
\[x^2-20x+20=0\]

Answer 10

you really do not want to take square roots of decimals or divide by decimals
use whole numbers

Answer 11

alright thank you!

Answer 12

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

Answer 13

And its not factorable right?