Meaning of "Solve over the reals?"

Question

I am a bit confused about the wording presented in some radical equation problems I have been seeing.

The directions state simply to "solve over the reals"

Here is an example:

$$\sqrt{x^2-26}=\sqrt{x-6}.$$

By squaring both sides and solving the resulting quadratic we get $x=-4, x=5$. Clearly both of these solutions are real, but when we evaluate the original equation at these values we get equality of two complex numbers.

My question is: is this wording implying that our solutions $x$ need to be real values, or that they need to be in the domain of each side of each expression in the equation to cause equality in the reals.

My interpretation is that "solve over the reals" means the solutions need to be real, but I may be wrong on this. If this is what is implied, is there a concise wording that would imply the need for equality in the reals as well?

Thank you!

Answer 1

Whenever you're told to solve an equation, you must always take into account the domains of the original equations.

Take for example the equation

$$\dfrac{x^2+5x+6}{x+2}=0\text{.}\tag{*}$$ Since division by zero is not valid, $x \neq -2$, but one could think to just multiply $x+2$ on both sides to get $$x^2 + 5x + 6 = 0$$ or $$(x+2)(x+3) = 0$$ yielding $x = -2$ or $x = -3$, but we must ignore $x = -2$ in this case, since it is not a solution to the original equation (*).

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In your example, you are told to solve the equation $$\sqrt{x^2-26}=\sqrt{x-6}$$ over the real numbers.

The domain of $\sqrt{x-6}$ is $[6, \infty)$.

Now note $$\sqrt{x^2-26}=\sqrt{(x-\sqrt{26})(x+\sqrt{26})}$$ and since $\sqrt{\cdot}$ is only defined for nonnegative ($\geq 0$) values, it must be the case that one of two things must happen:

1. Both $(x-\sqrt{26})\geq 0$ and $(x+\sqrt{26}) \geq 0$.
2. Both $(x-\sqrt{26})\leq 0$ and $(x+\sqrt{26}) \leq 0$.

Suppose (1) happens. This means that $$x - \sqrt{26} \geq 0 \implies x \geq \sqrt{26}$$ and $$x + \sqrt{26} \geq 0 \implies x \geq -\sqrt{26}$$ or in short, $x \geq \sqrt{26}$ (note the "and;" where $x \geq \sqrt{26}$ is where BOTH $x \geq \sqrt{26}$ and $x \geq -\sqrt{26}$. You can also think of this as intersecting number lines).

Suppose (2) happens. This means that $$x - \sqrt{26} \leq 0 \implies x \leq \sqrt{26}$$ and $$x + \sqrt{26} \leq 0 \implies x \leq -\sqrt{26}$$ or in short, $x \leq -\sqrt{26}$.

Thus, it must be the case that $x \in (-\infty, -\sqrt{26}] \cup [\sqrt{26}, \infty)$.

The domains of the two equations on both sides are $[6, \infty)$ and $(-\infty, -\sqrt{26}] \cup [\sqrt{26}, \infty)$. Since $\sqrt{26} \approx 5.09 < 6$, we will only be looking at the intersection of these two domains, which would be $[6, \infty)$.

If I solved $$\sqrt{x^2-26}=\sqrt{x-6}$$ I obtain $$x^2-x-20 = (x-5)(x+4) = 0$$ which yields $x = 5$ and $x = -4$, but neither of these solutions are in $[6, \infty)$. Thus, there are no solutions.

Answer 2

It looks like your problem is on the definition of the square root. In general the square root is a map $\sqrt{\cdot }\colon [0,+\infty) \rightarrow [0,+\infty) $ such that $\sqrt{x} = y$ if and only if $y \geq 0$ and $y^2 = x$.

If your variable is real (resp. complex) and you allow values in $\mathbb{C}$, i.e. $\sqrt{\cdot}\colon \mathbb{R} \rightarrow \mathbb{C}$ (resp. $\sqrt{\cdot}\colon \mathbb{C} \rightarrow \mathbb{C}$) you are using a different definition of the square root operator. You should check what your definition is in your reference.

Suppose that the definition is the first one above. Define $f(x) := \sqrt{x^2-26}$ and $g(x) := \sqrt{x-6}$. Since the square root operator takes non-negative real numbers in input and returns non-negative real numbers, the functions $f$ and $g$ are defined when $$ \begin{cases} x^2-26 \geq 0 \\ x-6 \geq 6. \end{cases} \Rightarrow \begin{cases} x \leq -\sqrt{26} \text{ or } x \geq \sqrt{26} \\ x \geq 6. \end{cases} $$ The solution of this system is $x \geq 6$. Now, since the square root operator is also bijective, you have that $f(x) = g(x)$ if and only if $f(x)^2 = g(x)^2$, which implies $(x-5)(x+4)=0$. So $x=5$ or $x=-4$. Both these roots are less than $6$, so they are not really solutions of the equation. You end up with no real solutions.