Question

find the solutions by completing the square?
Please help?!
.05x^2-x+1

Answer 1

it may not be possible but i just need help?!

Answer 2

@satellite73 can you help?

Answer 3

yes of course

Answer 4

if its not possible by completing the square im suppose to say why but im not sure where to even start here?

Answer 5

you want
\[.05x^2-x+1=0\] and you need to complete the square, so first step is to make the leading coefficient 1
since \(\frac{1}{.05}=\frac{100}{5}=20\) you need to multiply everything by 20 to accomplish this
\[20(.05x^2-x+1)=x^2-20x+20\]

Answer 6

it is always possible to complete the square

Answer 7

now we want to solve
\[x^2-20x+20=0\] first subtract 20 from both sides to get
\[x^2-20x=-20\] and then we "complete the square" by saying "half of 20 is 10 and \(10^2=100\) and write
\[(x-10)^2=-20+100\] i.e.
\[(x-10)^2=80\]

Answer 8

take the square root of both sides. don't forget the \(\pm\)

Answer 9

get
\[x-10=\pm\sqrt{80}\] and so
\[x=10\pm\sqrt{80}\]

Answer 10

since \(80=16\times 5\) you know \(\sqrt{80}=\sqrt{16}\times \sqrt{5}=6\sqrt{5}\) so you can also write the answer as
\[x=10\pm6\sqrt{5}\]

Answer 11

all steps are there, hope it is clear

Answer 12

Thank you