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Mathematics 12 Online
OpenStudy (amtran_bus):

I need help solving this rational equation

OpenStudy (amtran_bus):

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

OpenStudy (amtran_bus):

I know you square both sides

OpenStudy (amtran_bus):

So is it x+2 =X^2

OpenStudy (anonymous):

It's a radical equation, not rational equation, but yes, after squaring both sides, rewrite in standard form, then use quadratic formula.

OpenStudy (amtran_bus):

So \[X^2 -x-2?\]

OpenStudy (anonymous):

Yes, and it looks factorable too.

OpenStudy (rsadhvika):

*missing = 0

OpenStudy (rsadhvika):

and, \(x \ne 2\)

OpenStudy (anonymous):

? x=/=2 @rsadhvika

OpenStudy (rsadhvika):

sorry was a typo, it shud be \(x \ne -2\)

OpenStudy (amtran_bus):

Wolfram says the answer IS 2... http://www.wolframalpha.com/input/?i=find+solutions+to+sqrt+%28x%2B2%29+%3Dx

OpenStudy (anonymous):

2 is one of the solutions.

OpenStudy (rsadhvika):

ok it shud be \(x> -2\) :)

OpenStudy (anonymous):

I meant to the quadratic, If you try the other one in the original equation though . . .

OpenStudy (jdoe0001):

\(\bf \sqrt{x+2} =x \implies x + 2 = x^2 \implies 0 = x^2-x-2\\ \implies (x-\square?)(x+\square?)\)

OpenStudy (jdoe0001):

or \(\bf (x-\square?)(x+\square?)=0 \) for that matter

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