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is 2 √x+5+7=4 extraneous? will give medal
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\[\large\rm 2\sqrt{x+5}+7=4\]For these types of problems, try to solve for the square root.
Subtracting 7 from each side,\[\large\rm 2\sqrt{x+5}=-3\]Dividing by 2,\[\large\rm \sqrt{x+5}=-\frac32\]
Square root operation should always produce a `positive number`. There is no value that when we take the root of it, gives us a -3/2.
so is it?
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