Question

solve and determine any extraneous solutions.
2*(sqrt4-2x)=2-x

Answer 1

\[\large 2 \times (\sqrt4-2x)=2-x\]
or
\[\large 2 \times (\sqrt{4-2x})=2-x\]
??? @berrypicking

Answer 2

-6, 2 = x?
considering is (1-(x/2))^2 = (4-2x)

Answer 3

2nd equation 4-2x under radical sign

Answer 4

yes the answers are -6 and 2 but not sure how to set up problem. I know I have to square both sides but don't know what to do with the 2 in front of the radical sign

Answer 5

\[\large 2 \times (\sqrt{4-2x})=2-x\]
\[\large \sqrt{4-2x}=\frac {2-x}2\]
\[\large (\sqrt{4-2x})^2=(\frac {2-x}2)^2\]
\[\large 4-2x=\frac {x^2}4 -x +1\]
\[\large 3-x=\frac {x^2}4 \]
\[\large 12-4x={x^2} \]
\[\large 12-4x-x^2=0 \]
solve using quadratic method or factorize

Answer 6

as per previous answers: -6 or 2

Answer 7

how did you come up with x^2/4

Answer 8

sorry dude, went to sleep
\[\Large (\frac {2-x}2)^2\]
\[\Large (\frac 22 - \frac x2)^2\]
\[\Large (1 - \frac x2)^2\]
\[\Large (1 - \frac x2) (1 - \frac x2)\] now use FOIL
\[\Large (1*1 - 1*\frac x2 - 1*\frac x2 + (\frac x2)*(\frac x2)\]
\[\Large =\frac {x^2}4 -x +1\]