Is it Possible for an Absolute Value Equation to Equal a Negative Number?

[mathdad]

Precalculus by David Cohen 3rd Edition
Chapter 1, Section 1.2.

Question 68, page 11.

Before typing the textbook question, I must say that I have not been able to find a satisfactory answer to absolute value equations that equal a negative number.

Question:

Explain why there are no real numbers that satisfy the equation | x^2 + 4x | = - 12.

Aside from the question above, can we say the answer is a complex number?

 

[Ackbach]

Precalculus by David Cohen 3rd Edition
Chapter 1, Section 1.2.

Question 68, page 11.

Before typing the textbook question, I must say that I have not been able to find a satisfactory answer to absolute value equations that equal a negative number.

Question:

Explain why there are no real numbers that satisfy the equation | x^2 + 4x | = - 12.

Aside from the question above, can we say the answer is a complex number?

Actually, there won't be any complex solutions, either. That's because, if $z$ is a complex number, its magnitude $|z|$ is defined to be the real number representing its distance from the origin. So, if $z=a+ib$, then $|z|^2=z \cdot \bar{z} = (a+ib)(a-ib)=a^2+b^2,$ which is real and non-negative. Therefore its square root, $|z|,$ will be also.

Conclusion: the result of taking the magnitude (absolute value) of any quantity, real or complex, is always a real number. Moreover, that real number is always non-negative. There are no exceptions to this rule, either in the real numbers or in the complex numbers.

Think of it this way: you want to measure the distance from one thing to another thing. So you get out your tape measure and measure the distance. Can you ever get a negative number as the result? No? Well, you can always think of any absolute value/magnitude as the distance from one thing to another, even if one of them is zero: $|x|=|x-0|=$ distance from $x$ to $0$. You can also look at the definition of the absolute value function:
$$|x|=\begin{cases}\!\!\!\!&\phantom{-}x,\quad x\ge 0 \\ \!\!\!\!&-x,\quad x<0 \end{cases}.$$
The result in either case is always non-negative.

 

[mathdad]

What if there is a minus sign in front of the absolute value?

Let a be any real number.

- | a | = - a, right?

- | - a | = - (a) = - a, right?

 

[Ackbach]

What if there is a minus sign in front of the absolute value?

Let a be any real number.

- | a | = - a, right?

- | - a | = - (a) = - a, right?

Almost. You can think of the minus sign as canceling: $|a|=a$. But now, whether there's a solution or not depends greatly on $a$. If $a>0$, there is one solution: $a$. The reason why $-a$ is not a solution is that you'd have to plug it in on both sides of the equation, including the RHS. But then you have a magnitude sign equal to a negative number, which can't happen. The same goes for if $a=0$, but of course there, you already know what $a$ is! If $a<0$, there is again no solution, for the reasons we've stated above.

 

[mathdad]

Interesting ideas.