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OpenStudy (amtran_bus):
I need help solving this rational equation
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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 \[X^2 -x-2?\]
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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 \(x \ne -2\)
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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):
\(\bf \sqrt{x+2} =x \implies x + 2 = x^2 \implies 0 = x^2-x-2\\ \implies (x-\square?)(x+\square?)\)
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OpenStudy (jdoe0001):
or \(\bf (x-\square?)(x+\square?)=0 \) for that matter
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