Given the equation −4Square root of x minus 3 = 12, solve for x and identify if it is an extraneous solution. x = 0, solution is not extraneous x = 0, solution is extraneous x = 12, solution is not extraneous x = 12, solution is extraneous

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Given the equation −4Square root of x minus 3 = 12, solve for x and identify if it is an extraneous solution. x = 0, solution is not extraneous x = 0, solution is extraneous x = 12, solution is not extraneous x = 12, solution is extraneous

Mathematics
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ok so we have -\[-4\sqrt{x-3}=12\]divide both sides by -4 we get\[\sqrt{x-3}=-3\]square both sides u get -\[x-3=9\]add 3 to both sides u get \[x=12\] Plug in x=12 in the original equation i.e, \[-4\sqrt{x-3}=12\]we get \[-4\sqrt{12-3}=12\]simplifying we get-\[-4(3)=12\]\[-12=12\]which doesn't makes any sense so x=12 is an extraneous solution :)
Ahhhh, i see. Thanks soooo much!

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no problem :) u r welcome

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