Question

sqrt x + 2=x

Answer 1

x=x^2-4x+4...I'm kind of stuck here

Answer 2

Is the equation: \[\sqrt{x+2} = x\]

Answer 3

My apology, I lost internet connection. Yes, that's the equation.

Answer 4

square both sides in order to get rid of the radical:
x + 2 = x^2
x^2 - x - 2 = 0
x^2 - 2x + x - 2 = 0
x(x - 2) + 1(x - 2) = 0
(x-2)(x+1) = 0
x = -1 or 2.
put back each solution in the original equation and see if it is valid:
x = -1, sqrt(-1 + 2) = sqrt(1) = 1 which is not equal to -1. So this is not a valid solution.
Try x = 2: sqrt(2 + 2) = sqrt(4) = 2 which is the right hand side. So this solution is good.
x = 2.

Answer 5

Ok, I need a bit more help with this problem. I worked it out differently and got x=1 as the reject and x=4; {4} \[x=x ^{2}-4x=4\]

Answer 6

\[0=x ^{2}-5x+4\]

Answer 7

How did you get the above equation from the given equation sqrt (x + 2) = x ?

Answer 8

\[0=(x-1) x=1 (x-4) x=4\]

Answer 9

How did you get x = x^2 - 4x + 4 from the given equation sqrt (x + 2) = x ?

Answer 10

Square root is different from squaring.

Answer 11

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

Answer 12

Look at my first reply. That is the given problem. How did you go from there to your last reply?

Answer 13

oh, maybe I wrote the problem incorrectly...would it make a difference if the equation was \[\sqrt{x} + 2 = x\]

Answer 14

Is the given problem \[\sqrt{x} + 2 = x \quad or \quad \sqrt{x+2} = x \quad ?\]

Answer 15

Yes that would make a difference.

Answer 16

My apology it's \[\sqrt{x} + 2 = x\]

Answer 17

\[\sqrt{x} + 2 = x\\\sqrt{x} = x - 2\\\sqrt{x})^{2} = (x-2)^{2}\\x = x^2 - 4x + 4\\x^2-5x+4=0\\x^2-4x-x+4 = 0\\x(x-4)-1(x-4)=0\\(x-1)(x-4)=0\\x=1 ~or~x=4\]Test each solution in the original equation.

Answer 18

x = 1 gives sqrt(1) + 2 = 1 which is not valid.
x = 4 gives sqrt(4) + 2 = 2 which is valid.

Therefore, the solution is x = 4.

Answer 19

Ok, I see. Thank you very much for your help! :-)

Answer 20

You are welcome.