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Mathematics 9 Online
OpenStudy (anonymous):

Solve square root of x plus 2 end root minus 2 equals x

OpenStudy (jonnyvonny):

\[\sqrt{x+2}-2=x\]?

OpenStudy (agent0smith):

Use symbols not words.

OpenStudy (anonymous):

best i can do is sqrt(x+2) -2 = x

OpenStudy (anonymous):

@JonnyVonny Yes!!

OpenStudy (agent0smith):

^that "sqrt(x+2) - 2 = x" is perfectly fine, easy to read. First add 2 to both sides Then, squrare both sides

OpenStudy (anonymous):

sqrt(x+2) = 2x?

OpenStudy (agent0smith):

What is x + 2 equal to? x+2. Not 2x, you can't combine unlike terms.

OpenStudy (anonymous):

Ah right I see, I never really succeeded at math >_> x + 2, subract 2 from the the x + 2?

OpenStudy (agent0smith):

See the steps I posted above, there was only two.

OpenStudy (anonymous):

x=1

OpenStudy (agent0smith):

x=1 will not work in this equation. And show your work. sqrt(x+2) - 2 = x after adding 2 to both sides sqrt(x+2) = x + 2 now square both sides.

OpenStudy (anonymous):

squaring sqrt(x+2) would be sqrt(x+2)^2 ?

OpenStudy (anonymous):

would it be no solution

OpenStudy (jonnyvonny):

Well, you have:\[\sqrt{2x+2}=2+x \rightarrow 2x+2=(2+x)^2\] From then, you factor out the square, and get x to one side, then solve.

OpenStudy (anonymous):

(2x+1)(2x-1) ?

OpenStudy (agent0smith):

Nope... you seem to be getting things very mixed up.

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