MHB Solve \[\sqrt{30-2y} + 3 = y\] - Verify Solutions & Find Out What Happened

[eleventhxhour]

6) Solve the equation \[\sqrt{30-2y} + 3 = y\] Is/are the solution(s) in the domain of both expressions? Verify your solution(s). What happened?

So, I'm not really sure how to start this because of the square root sign...I guess you'd move the y over first, but then I'm not sure what to do next.

Thanks!

 

[Ackbach]

6) Solve the equation \[\sqrt{30-2y} + 3 = y\] Is/are the solution(s) in the domain of both expressions? Verify your solution(s). What happened?

So, I'm not really sure how to start this because of the square root sign...I guess you'd move the y over first, but then I'm not sure what to do next.

Thanks!

I would try $\sqrt{30-2y}=y-3$, and then square both sides. What happens then?

 

[eleventhxhour]

I would try $\sqrt{30-2y}=y-3$, and then square both sides. What happens then?

\[\sqrt{30-2y} = \sqrt{y-3}\]
\[\sqrt{30+3} = \sqrt{y+2y}\]
\[\sqrt{33} = \sqrt{3y}\]

Is this correct so far? I'm not sure what to do next, though.

 

[Ackbach]

\[\sqrt{30-2y} = \sqrt{y-3}\]
\[\sqrt{30+3} = \sqrt{y+2y}\]
\[\sqrt{33} = \sqrt{3y}\]

Is this correct so far? I'm not sure what to do next, though.

I'm afraid your steps there are not correct. You cannot distribute addition or subtraction over the square root function. Multiplication is about the only mathematical operation that you can distribute over addition or subtraction! Moreover, you somehow got a square root on the RHS where there wasn't one before, and while doing nothing to the LHS. Remember the Golden Rule of Algebra: what thou doest to one side of the equation, thou must do to the other.

Here's the next step:
$$30-2y=(y-3)^{2}.$$
Can you continue?

 

[eleventhxhour]

I'm afraid your steps there are not correct. You cannot distribute addition or subtraction over the square root function. Multiplication is about the only mathematical operation that you can distribute over addition or subtraction! Moreover, you somehow got a square root on the RHS where there wasn't one before, and while doing nothing to the LHS. Remember the Golden Rule of Algebra: what thou doest to one side of the equation, thou must do to the other.

Here's the next step:
$$30-2y=(y-3)^{2}.$$
Can you continue?

$$30-2y = (y-3)(y-3)$$
$$30-2y = y^2-6y+9$$
$$30-9 = y^2-6y+2y$$
$$0 = y^2 - 4y - 21$$
$$0 = (y+3)(y-7)$$
$$y = -3, y = 7$$

Is this correct?
Also, I don't understand what it means by "Is/are the solution(s) in the domain of both expressions?"

 

[Prove It]

$$30-2y = (y-3)(y-3)$$
$$30-2y = y^2-6y+9$$
$$30-9 = y^2-6y+2y$$
$$0 = y^2 - 4y - 21$$
$$0 = (y+3)(y-7)$$
$$y = -3, y = 7$$

Is this correct?
Also, I don't understand what it means by "Is/are the solution(s) in the domain of both expressions?"

Look at your original equation. It as a square root in it. You can never have the square root of a negative number. So what values of y aren't possible?

 

[Ackbach]

Look at your original equation. It as a square root in it. You can never have the square root of a negative number. So what values of y aren't possible?

Just to reinforce Prove It's excellent suggestion: you should ALWAYS check your final solutions against the original equation to see if they make sense. Make that a habit.

 

[eleventhxhour]

Just to reinforce Prove It's excellent suggestion: you should ALWAYS check your final solutions against the original equation to see if they make sense. Make that a habit.

So is the value y = -3 not possible?

\[\sqrt{30-2(-3)} + 3 = -3\]
\[9 ≠ -3\]

And so does that mean that negative numbers aren't in the domain of the expression?

 

[Ackbach]

So is the value y = -3 not possible?

\[\sqrt{30-2(-3)} + 3 = -3\]
\[9 ≠ -3\]

And so does that mean that negative numbers aren't in the domain of the expression?

This is an interesting problem; it could be sometimes that squaring both sides of an expression produces extraneous solutions that are not in the domain of a square root. That's one thing you have to check. In your case, however, squaring both sides of an equation produces an extraneous solution, period. The extraneous solution does not happen to violate the domain rules of the square root, but it's still simply not a solution of the original equation.

The original question is a bit ambiguous, since the term "domain" applies to functions, and not equations. So the answer is that $y=7$ is the only solution, although both solutions are in the domain of both sides of the equation, each side viewed as a function.