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OpenStudy (anonymous):

x+2square root of x -3=0 solve for x

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OpenStudy (anonymous):

wait do u mean x+2 * sqrt of x-3 ???

OpenStudy (anonymous):

x + 2sqrt(x-3) = 0 is this correct?

OpenStudy (anonymous):

No its \[x+2\sqrt{x}-3=0\]solve for x

OpenStudy (dumbcow):

in that case, treat it like a quadratic. u = sqrt(x) -> u^2+2u-3 = 0

OpenStudy (anonymous):

2sqrt x = 3-x square both sides 4x = 9-6x+x^2 x^2 -10x + 9 = 0 (x-9)(x-1) = 0 x = 9 or 1 x = 1 satisfies the original equation and is the only solution

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OpenStudy (anonymous):

notice x=9 does not satisfy the equation

OpenStudy (anonymous):

this happens sometimes when you solve equations containing radicals by squaring - always check your results by substitution

OpenStudy (anonymous):

ok thank you

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