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For some reason, I'm forgetting how to work with square roots ... I know, not a great look. Anyway, I'm trying to solve for $x$.

$$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$

Since I see a square root, I divide the domain into two cases: (1) $x > 0$ and (2) $x < 0$.

(1)

$$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$ $$\frac{10x}{\sqrt{x^2}} = 10$$

Here is where I get confused, do I consider $\sqrt{x^2} = x $ or $-x$?

(2) Similarly, $$\frac{10(-x)}{\sqrt{10(-x)^2}} = \sqrt{10}$$ $$\frac{10(-x)}{\sqrt{(-x)^2}} = 10$$

Blue
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beginner
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    $\sqrt {x^2}=|x|$, not $x$, so the sign of $x$ needs to be taken into account. – lulu Jul 14 '21 at 21:50
  • How does $\sqrt{x^2} = \vert x \vert$? – beginner Jul 14 '21 at 21:55
  • It's effectively defined that way. The square root function always takes the positive root. The point of splitting it up into two cases is so you know whether the positive root of $\sqrt{x^2}$ is $x$ ($x>0$) or $-x$ ($x<0$). – Amaan M Jul 14 '21 at 21:56
  • wikipedia- Square root – Asher2211 Jul 14 '21 at 21:56
  • Okay so 4 and −4 are square roots of 16, because 4^2 = (−4)^2 = 16. But the square root $function$ always takes the positive root? – beginner Jul 14 '21 at 21:57
  • @beginner That is the definition of the square root. It always takes values $≥0$. – lulu Jul 14 '21 at 21:59
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    I wouldn't even think of 4 and -4 as square roots of 16, 4 is the square root of 16. The confusion often comes from the fact that both 4 and -4 are solutions to the equation $x^2 = 16$. It's tricky because people often think of $x^2$ and $\sqrt{x}$ as inverses, but that only holds for $x\ge 0$. – Amaan M Jul 14 '21 at 21:59
  • Ohhh. Okay. I didn't know that. I have some vague memory from ninth grade of the teacher writing $\sqrt{4} = 2,-2$ on the board, and went off that. – beginner Jul 14 '21 at 22:03
  • Not sure if this is important but why is it defined to be non-negative? How does this work out in terms of arithmetics? – beginner Jul 14 '21 at 22:07
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    Yeah this is a pretty common sticking point. Saying $4$ and $-4$ are both square roots is valid, but the square root function is a function so it can only give one output for any given input, so it gives the positive value, also called the principal value. So $4$ and $-4$ are both square roots of $16$ but $\sqrt{16}$ is uniquely $4.$ This same issue also appears with inverse trig functions because there are infinitely many angles that will have the same sine or cosine, so we restrict the range of the inverse so we uniquely define one. – Stephen Donovan Jul 14 '21 at 22:08
  • There is a distinction between the square roots of $16$, which are the two numbers $4$ and $-4$, and the square root of $16$, which is the single number $\sqrt{16}=4$. The square root symbol always signifies the singular root, never the pair of roots. – Arthur Jul 14 '21 at 23:39

4 Answers4

2

$$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$

First of all, if $x=0$ then the left-hand side $\frac00$ is undefined; so we reject that as a solution, and assume $x\ne 0$. So $\sqrt{10x^2}$ is positive, and we can multiply both sides by it to get $$10x=\sqrt{10}\sqrt{10x^2}=10\sqrt{x^2}$$ Dividing by $10$ gives $$x=\sqrt{x^2}$$ And we know that $\sqrt{x^2}=|x|$, so we get $$x=|x|$$ You can take it from here.

TonyK
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Take $x$ and square it. You will get a non-negative number, independently of whether $x$ is positive or negative.

Now, calculate the square root of the result. You will get a non-negative number. Notice that when it comes to real numbers, you are not allowed to take the square root of a negative number.

When working with non-negative real numbers, taking the square root and squaring are operations that cancel out. Thus for all $x >= 0, \space \sqrt(x^{2}) =x.$

But if $x$ is smaller than $0$, squaring it turns it into a positive number. Then one is allowed to take the square root. But the square root only gives non-negative outputs. Thus for all $x < 0, \space \sqrt(x^{2}) = -x.$ Hence for all real $x:\sqrt(x^{2}) = |x|.$

As you can see, the process results in "forgetting" the sign of the number one is working with.

Artur Wiadrowski
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Your way is perfectly fine up here

$$\frac{10x}{\sqrt{x^2}} = 10 \iff \frac{x}{\sqrt{x^2}} = 1$$

with $x\neq 0$, now we can use that $\sqrt{x^2}=|x|$ indeed

  • $x>0 \implies \sqrt{x^2}=x=|x|$
  • $x<0 \implies \sqrt{x^2}=-x=|x|$
user
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  1. None of the other Answers have pointed out your most remarkable error:

    $$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$ I divide the domain into two cases:   (1) $x > 0,\;$ and (2) $x < 0$. $$\text{Case (2)}\quad \frac{10(-x)}{\sqrt{10(-x)^2}} = \sqrt{10}$$

    This is illogical and wrong: writing $$-x$$ is not taking the case in which $x$ is negative—it is merely inverting the sign of $x.$

    Think of the minus sign not as an adjective, but as an arithmetic operator; read $-x$ not as “negative $x$”, but as “minus $x$”.

    In other words, $-x$ means

    • neither that $x$ is negative ($x$ can still have positive value $7$ so that minus $x$ is negative),
    • nor that $-x$ is negative ($-x$ has positive value $7$ when $x=-7$).

  2. While it is true that both $-2$ and $2$ are square roots of $4$—just as $4$ has seven $7^\text{th}$ roots, and so on—the issue is that $$\sqrt[n] x$$ (the radical sign) has a very particular and clear meaning: for nonnegative real $x, \;\sqrt[n] x$ is defined as the principal $n^\text{th}$ root of $x,$ i.e., $$\sqrt[n] {y^n}=|y|.$$

  3. However, when $x$ is negative or has an imaginary component, due to ambiguity, I avoid using the radical sign altogether: for example, it is unclear whether $$\sqrt[3] {-1}$$ means $$-1\quad\text{(real)}$$ or $$e^{i \frac\pi3}\quad\text{(smallest nonnegative argument)}.$$

ryang
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