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

[RChristenk]

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?

 

[PeroK]

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.
$$y =1 \implies y^2 =1 \implies y = \pm 1$$

 

[FactChecker]

Good catch! So many students neglect to check their answers back in the original equations. There are a few things that can introduce false solutions, so it is always wise to check them.

 

[PeroK]

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.

 

[SammyS]

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

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 n't work.

 

[MatinSAR]

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.

 

[Mark44]

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.

 

[WWGD]

For comparison, in Linear Algebra, there are operations you can do on your equations, system, that preserve the solutions, while others don't. In this setting, squaring both sides introduces a solution to your equation.

 

[SammyS]

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 $y=9$.