## fluffyunicorn 3 years ago how do i factor y=0.05x^2-x+1

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1. satellite73

you cannot, at least you cannot using rational numbers

2. fluffyunicorn

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

3. Monkeyball

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

4. Monkeyball

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

5. satellite73

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

6. satellite73

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

7. Monkeyball

so I did it wrong?

8. fluffyunicorn

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

9. satellite73

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

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

11. fluffyunicorn

alright thank you!

12. satellite73

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

And its not factorable right?