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Find the solution of 3 square root x+5=-9, and determine if it is an extraneous solution. The options are: X=4; not extraneous X=4; extraneous X=0; extraneous X=0; not extraneous
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so... ok.. what do you get for "x" from \(\bf 3\sqrt{x+5}=-9\)?
X=4; extraneous ?
@jdoe0001
is it? what makes you think so?
well, notice \(\bf 3\sqrt{{\color{brown}{ x}}+5}=-9\implies 3\sqrt{{\color{brown}{ 4}}+5}=-9\implies 3\cdot \sqrt{9}=-9 \\ \quad \\ 3\cdot 3\ne -9\) thus, if x = 4, it IS extraneous
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even though x = 4 is correct, procedure wise
I subtracted 5 which got 4. But how do you determine if it is extraneous?
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