How do you solve $y = \sqrt{12 - y}$ $y=sqrt(12-y)$ and check your solution?

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Answer 1 · 2017-04-20T20:54:26

See below.

Let's start by squaring both sides.

$y^{2} = 12 - y$ $y^2=12-y$

$y^{2} + y - 12 = 0$ $y^2+y-12=0$

Factoring,

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

So $y = - 4, 3$ $y=-4,3$

We can check to see which of these solutions work.

If $y = - 4$ $y=-4$ , then:

$- 4 \neq \sqrt{12 - (- 4)}$ $-4\nesqrt(12-(-4))$ , as a square root does not come out to be negative (in real numbers).

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

Now let's try $y = 3$ $y=3$

$3 = \sqrt{12 - 3} = \sqrt{9} = 3$ $3=sqrt(12-3)=sqrt9=3$ is true.

The only solution is therefore $y = 3$ $y=3$ .

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