Question

Please help! Solve : (equation below)
Hint 2x-x^2 = 1 - (1-2x+x^2)

Answer 1

\[\sqrt{2x-x ^{2}}=(x-1)^{2}\]

Answer 2

you can see what things become by getting rid of that radical
you can do that by square both sides

Answer 3

Then you get \[2x-x ^{2}=x ^{4}-4x ^{3}+6x ^{2}-4x+1\] and I don't know what to do from there.

Answer 4

you could get everything on one side
but usually a hint suggest easier way
so did you try to use your hint yet?

Answer 5

\[\sqrt{1-(x-1)^2}=(x-1)^2 \]

Answer 6

now square both sides

Answer 7

should look tons prettier

Answer 8

\[\sqrt{2x-x^2}=(x-1)^2 \\ \sqrt{1-(x^2-2x+1)}=(x-1)^2 \\ \sqrt{1-(x-1)^2}=(x-1)^2 \\ \text{ \square both sides } \\ 1-(x-1)^2=(x-1)^4 \\ 0=(x-1)^4+(x-1)^2-1 \]
all this is is a quadratic in terms of (x-1)^2

Answer 9

recall you can solve a quadratic equation:
\[a(u(x))^2+b(u(x))+c=0 \text{ like this } \\ u(x)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
and yes u(x) is just some function u of x

Answer 10

\[\text{ you have } [(x-1)^2]^2+[(x-1)^2]-1=0\]
I should let you finish

Answer 11

Thanks I wasn't replying because I was just figuring out the problem.

Answer 12

You've been very helpful so thank you so much!

Answer 13

np
that is actually a nice algebra question