Why Does Solving y+3=3√(y+7) Yield Extraneous Solutions? (2024)

##y+3=3\sqrt{y+7}##

Square both sides:

##\Rightarrow y^2+6y+9=9y+63##

##\Rightarrow y^2-3y-54=0##

##\Rightarrow (y-9)(y+6)=0##

##y=9, -6##

But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?

RChristenk said:

Homework Statement: Solve ##y+3=3\sqrt{y+7}##
Relevant Equations: Basic algebra

##y+3=3\sqrt{y+7}##

Square both sides:

##\Rightarrow y^2+6y+9=9y+63##

##\Rightarrow y^2-3y-54=0##

##\Rightarrow (y-9)(y+6)=0##

##y=9, -6##

But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?

Because it's not a solution. When you square an equation you may introduce an addition solution. E.g.
$$y =1 \implies y^2 =1 \implies y = \pm 1$$

PeroK said:

Because it's not a solution. When you square an equation you may introduce an addition solution. E.g.
$$y =1 \implies y^2 =1 \implies y = \pm 1$$

Actually, better is:
$$y =1 \implies y^2 =1 \Leftrightarrow y = \pm 1$$And we see that the first step is not reversible.

To further illustrate @PeroK's point:
If your problem had been the following then ##y=-3## would had been the solution.

RChristenk said:

Solve ##y+3=-3\sqrt{y+7}##

Square both sides:

##\Rightarrow y^2+6y+9=9y+63##
. . .

##y=9, -6##

If you plug in ##y=-6## into the original equation, you get ##-3=-3## . So it does

work.

RChristenk said:

But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?

Because it is a solution to ## y^2+6y+9=9y+63##. This is not your original equation. So you should check all answers of ## y^2+6y+9=9y+63## to see if they work for original question or not.

RChristenk said:

Solve ##y+3=3\sqrt{y+7}##
<snip>
##y=9, -6##
But if you plug in ##y=-6## into the original equation, you get ##-3=3## . So it doesn't work. Why?

Something that hasn't been mentioned so far is that it's good practice to take a few moments before you start doing any manipulations. The right side of the first equation must be greater than or equal to zero, hence, it must be true that ##y + 3 \ge 0## which implies that ##y \ge -3##. This would rule out the negative solution you found.

As @SammyS noted, if the equation had been ##y+3=-3\sqrt{y+7}##, here the right side must be less than or equal to zero, hence we must have ##y + 3 \le 0## or ##y \le -3##. For this scenario, we would have to discard y = 9 as an extraneous solution.

RChristenk said:

Solve:

##y+3=3\sqrt{y+7}##

Here's another method which can help you avoid dealing extraneous solutions.

Write the equation as a quadratic equation in ##\sqrt{y+7\,}\ ## .

##\displaystyle \quad y+3=3\sqrt{y+7\,}##

##\displaystyle \quad y+7=3\sqrt{y+7\,}+4##

##\displaystyle \quad \left(\sqrt{y+7\,}\right)^2-3\sqrt{y+7\,}-4=0##

Factoring gives:

##\displaystyle \quad \left(\sqrt{y+7\,} -4\right) \left(\sqrt{y+7\,}+1\right)=0##

Only one of these tactors can be zero. That is ##\displaystyle \ \left(\sqrt{y+7\,} -4\right)=0\ ## giving ##y=9##.