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Find the solution of 3 square root of x+5=-9 these are the anser choices: x = 4; not extraneous x = 4; extraneous x = 0; extraneous x = 0; not extraneous
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can you write the q in a better way perhaps?
\[\sqrt[3]{2}\]
my bad hold up
\[\sqrt[3]{x2}\]
omgsorry i messed up
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hold up i got it
\[\LARGE 3 \sqrt {x+5} = -9\] ?
3\[3\sqrt{x+5=-9}\]
yeah thats it! sorry im new to this
the question is wrong..sqrt always gives a +ve number..how can RHS be negative still..square both the sides \[\LARGE 9 (x+5) = 81 = x+5 = 9 => x=4\] but that doesn't satify the equation..it does if RHS = + 9
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thats what the question said
x=4,extraneous is the answer
ok thank veryy much
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