SOLUTION: how to solve the square root of y+3 = y-3

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Question 138197This question is from textbook Intermediate Algebra
: how to solve the square root of y+3 = y-3 This question is from textbook Intermediate Algebra

Answer by solver91311(24713) About Me  (Show Source):
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sqrt%28y%2B3%29=y-3

Square both sides (since we are squaring the equation, we have to remember to check for extraneous roots at the end)
y%2B3=%28y-3%29%5E2
y%2B3=y%5E2-6y%2B9
y%5E2-6y-y%2B9-3=0
y%5E2-7y%2B6=0

The quadratic factors because -1%2A-6=6 and -1%2B%28-6%29=-7

%28y-6%29%28y-1%29=0

Using the zero product rule:

y-6=0 => y=6

or

y-1=0 => y=1

Check for extraneous roots:
sqrt%286%2B3%29=6-3
sqrt%289%29=3 True

sqrt%281%2B3%29=1-3
sqrt%284%29%3C%3E-2 (radical sign means the positive square root by convention)
So 1 is an extraneous root.

The solution set is {6}