How do you solve #sqrt(y+21)-1=sqrt(y+12)#?

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Answer 1 · 2015-10-17T16:14:11

The solution is #y=4#.

Square both terms:

#y+21 - 2sqrt(y+21) + 1 = y+12#.

Isolate the root:

#y+21 - y - 12 +1 = 2sqrt(y+21)#, which simplifies into

#2sqrt(y+21) = 10#

Square both terms again:

#4(y+21)=100 \iff 4y+84 = 100 \iff 4y=16#.

Solving by #y#, we get #y=4#.

This solution is acceptable, since both the original square roots are well defined in that point.