How do you solve #y=sqrt(12-y)# and check your solution?

Question

2479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-20T20:54:26

See below.

Let's start by squaring both sides.

#y^2=12-y#

#y^2+y-12=0#

Factoring,

#(y+4)(y-3)=0#

So #y=-4,3#

We can check to see which of these solutions work.

If #y=-4#, then:

#-4\nesqrt(12-(-4))#, as a square root does not come out to be negative (in real numbers).

Thus, #y=-4# is extraneous, and is not a solution.

Now let's try #y=3#

#3=sqrt(12-3)=sqrt9=3# is true.

The only solution is therefore #y=3#.