Question

Find the solutions of the function by completing the square. If this method cannot be used explain why.
I already did it I need help.

Answer 1

y = 0.05x^2 -x +1

20y=x^2 - 20x + 20

20y-20=x^2-20x

20y+80= x^2 - 20x + 100

20y+80=(x - 10)^2

20y=(x-10)^2-80

y=(0.05)(x - 10)^2-4

Answer 2

I completed the square what does it want from me?

Answer 3

jdoe?

Answer 4

well, finding the solutions means, finding the "roots" or zeros of the quadratic equation

Answer 5

so you'd set y = 0 and solve for "x", to get the 2 roots

Answer 6

ok...

Answer 7

\(\bf y=(0.05)(x - 10)^2-4 \implies y = \cfrac{5}{100}(x - 10)^2-4\\
\textit{setting y = 0}\\
0 = \cfrac{5}{100}(x - 10)^2-4 \implies 4= \cfrac{5}{100}(x - 10)^2\\
80=(x - 10)^2 \implies \pm\sqrt{80} = x-10\\
---------------------\\
80 = 2 \times 2 \times 2\times 2\times 5\implies 2^4\times 5 \implies (2^2)^2 \times 5\\
---------------------\\
\pm\sqrt{80} = x-10 \implies \pm\sqrt{(2^2)^2 \times 5} = x-10 \implies \pm4\sqrt{5}= x-10\)

Answer 8

holy crap!

Answer 9

and surely you can get it from there

Answer 10

uhhh

Answer 11

so it cannot be used?

Answer 12

Im so confused

Answer 13

well, what would be the values for "x" from there?

Answer 14

the last line you have

Answer 15

the last line isn't simplified yet

Answer 16

i don't know :(

Answer 17

well, anyhow, the completion was fine btw... so
\({\bf 4\sqrt{5}= x-10 \implies}
\begin{cases}+4\sqrt{5}+10=& x\\
-4\sqrt{5}+10=& x\\
\end{cases}\)

Answer 18

that's all it's asking??? lol...

Answer 19

yeah, it's asking for the "zeros" or "roots", which would be the "solutions"
so using your square completion, we did find them

Answer 20

ohh thank you so much

Answer 21

yw