Question

solve: sqrt{7-2x}=x-2

Answer 1

Answer is 3

Answer 2

To get 3

Answer 3

what is the work for it?

Answer 4

You take the sqrt of both sides

Answer 5

You will be left with 7-2x= x^2 -4

Answer 6

Next combine like terms

Answer 7

x^2-2x-11

Answer 8

x^2+2x-11

Answer 9

I made a mistake somewhere

Answer 10

Let me check

Answer 11

I should really do these things on paper

Answer 12

@Microrobot , please type more than 1 step in each post
and you made a mistake squaring right side

Answer 13

I did the same thing before too

Answer 14

@Microrobot You don't seem to be getting the idea. DO NOT just dump answers. The fact that the OP had to ask you to show your work exposes the pathology. The Code of conduct is important.

Answer 15

Thank you dumbcow

Answer 16

I am explaining the answer @tkhunny

Answer 17

is it positive 4 instead of negative

Answer 18

I made a mistake somewhere

Answer 19

So im doing it on paper to prevent errors

Answer 20

and then reexplain

Answer 21

\(\sqrt{7-2x} = x-2\)

First, observe that \(7 - 2x \ge 0\) or \(2x \le 7\) or \(x \le 7/2\) WHY?

Second, observe that \(x - 2 \ge 0\) or \(x \ge 2\) WHY?

Answer 22

@Microrobot Your first two posts were just answers. EIGHT of your other posts are just chatter. Just take a little more time and be sure of what you are doing. So far, you're quite a bit more on the confusing side of things.

Answer 23

Ok you can take this one

Answer 24

Sure

Answer 25

Sorry about that, thank you for informing me

Answer 26

@sydpendleton ,u still need any help?

Answer 27

yes please

Answer 28

ok
@tkhunny showed you the domain restrictions which tells you the solution must be in interval 2<x<7/2

@Microrobot , was on right track...square both sides, then solve the quadratic
\[(\sqrt{7-2x})^{2} = (x-2)^{2}\]
\[7-2x = x^{2} -4x+4\]
\[x^{2} -2x-3 = 0\]
\[(x-3)(x+1) = 0\]
only 1 solution in domain....x=3

Answer 29

Thanks!

Answer 30

syd badge dumbcow

Answer 31

thank you!

Answer 32

Select Best Response for dumbcow

Answer 33

Thanks!

Answer 34

I did :)

Answer 35

Thanks!