5
$\begingroup$

A few months ago I asked a question about the $\pm$ symbol because I was confused about it... I still carry the same confusion (which really bugs me) but I think the real confusion has to do with the square root and principal square root. I hope I can finally grasp the concept with the following two questions...

Question 1:

$\sqrt{12x^2} = \sqrt{4\cdot3x^2} = 2|x|\sqrt{3}$

Why do we only have to keep in mind that $x$ could be negative, yet we just factorize $\sqrt{12}$ to $\sqrt{3\cdot4}$ and put the 2 right in front of the radical sign?

$\sqrt{12}$ could be factorized as $\sqrt{-3 \cdot -4}$ as well.

Question 2:

I've been told to take the square root of both sides in the following equation, but the square root doesn't have its own symbol right? Only the principal root does... So if you'd use the radical sign you're using the principal root and therefore you're missing out on a solution in the following scenario:

$(x + 8)² = 1 \iff \sqrt{(x + 8)^2} = \sqrt1 \iff x + 8 = 1 \iff x = -7$

Does that mean that whenever we want to take the square root of something (not the principal) we just use a $\pm$ symbol instead? For example:

$(x + 8)^2 = 1 \iff x + 8 = \pm 1 \iff x = -8\pm1$

$\endgroup$
3
  • 1
    $\begingroup$ You've made a mistake: $\sqrt{{(x+8)}^2}=|x+8|$ since it is the positive root, so everything should work if you change that (for Q2). $\endgroup$ – H. Kissos Apr 22 '18 at 19:18
  • $\begingroup$ @user1534664 $x+8=\pm1 \iff x=8\pm1$ Nope. Should be $x+8=-8\pm1$. $\endgroup$ – L. F. Feb 5 '19 at 2:56
  • $\begingroup$ Yes $\sqrt{12} = \sqrt{-3*-4}$ that that doesn't helps us do anything. Before we can say "hey, we can split $12$ into two parts and take the square roots of each part we have to verify we can do that. Suppose we have $K= a\sqrt b$ and $K,a,b>0$. then $K^2 =a^2b$ so $K = \sqrt{a^2b}$ so $a\sqrt b=\sqrt{a^2b}$. So we verified we can split. but $\sqrt{-3*(-4)}= \sqrt{-3}\sqrt{-4}$ wont help us as those are not defined for the real numbers. $\endgroup$ – fleablood Jun 14 '20 at 2:25
2
$\begingroup$

For question 1, you are correct that you could factor $12=-3 \cdot -4$, but it is not useful. You are choosing to use $12=3 \cdot 4$ because that is useful. You know $3,4 \gt 0$ because you chose them that way. $x$ is a variable, and we don't know its sign, so we need the absolute value signs. We pull out the $2$ because we like $2 \sqrt 3$ better than $\sqrt {12}$. It is a matter of taste, and in some intermediate calculations we would prefer $\sqrt {12}$ or some other form.

For question 2, yes I would just go to $x+8 = \pm 1$

$\endgroup$
2
  • $\begingroup$ Q2: So just to confirm, we don't have a symbol for square roots that give negative answers as well? So skipping that square root step is applicable to all situations? $\endgroup$ – user1534664 Oct 25 '14 at 14:40
  • 1
    $\begingroup$ That's right. In some cases it is useful to put the $\pm$ sign in front of the square root $\endgroup$ – Ross Millikan Oct 25 '14 at 14:49
1
$\begingroup$

Q1: If $a,b$ are nonnegative, then $\sqrt{ab}=\sqrt a\sqrt b$. If $a<0$ or $b<0$ (and you are not into complex numbers), you better not talk about the square roots on the right.

Q2: In general $a=b\implies f(a)=f(b)$. Since $(x+8)^2$ and $1$ are both nonnegative, we can ltake $\sqrt{\ }$ for the function $f$ and obtain $$(x+1)^2=1\implies \sqrt{(x+1)^2}=\sqrt 1.$$ (Actually, since the square root function is injective on $[0,\infty)$, the use of "$\iff$" would be justifed, but itis not for general $f$). Now recall that $$\tag0\sqrt {t^2}=|t|$$ for all real $t$ (wheras $\sqrt{t^2}=t$ only for nonnegative $t$, and we don't know if $x+1$ is nonnegative). Hence by simplificatiopn of expressions we find $$\sqrt{(x+1)^2}=\sqrt 1\iff |x+1|=1$$ Now the appearence of $|\cdot|$ suggests that we investigate two cases: $$\tag a x+a\ge 0\quad\land\quad x+1=1$$ $$\tag b x+a< 0\quad\land\quad -(x+1)=1$$ So the $\pm$ sign (or any other distinction of two cases, e.g. as I just did) ultimately comes from equation $(0)$.


An alternative way to proceed from the beginning is as follows, using the third binomial formula: $$ (x+1)^2=1\iff (x+1)^2-1=0\iff((x++1)+1)((x+1)-1)=0$$ and now use that a product is zero if and only if at least one of the factors is zero.

$\endgroup$
1
  • $\begingroup$ What do you mean by "square roots on the right"? $\endgroup$ – user1534664 Oct 25 '14 at 14:42
0
$\begingroup$

For question 2, you could disambiguate the situation by factoring. Remember that the plus-or-minus symbol is just a convenience, and, like most conveniences, it involves some loss of something, clarity in this case. Nothing is free, and the cost of convenience is loss of clarity, or safety (as in trying to cross the track before the oncoming train passes), or something else.

$\endgroup$
1
  • $\begingroup$ There isn’t a need to factor the equation in question 2. If you know what the plus-or-minus symbol means, then it doesn’t involve a loss of clarity (just like any other symbol). $\endgroup$ – Radial Arm Saw Jun 14 '20 at 1:40
-1
$\begingroup$
  1. We have to keep in mind that $x$ might be negative because we don't know what $x$ is and it might by negative.

But we don't have to keep anything in mind about $12$ because we know exactly what $12$ is.

But perhaps we need to understand why we are allowed to split $\sqrt {ab}=\sqrt{a}\sqrt{b}$ in the first place.

Real Number Answer (Assumes Since $m^2 \ge 0$ for all $m \in \mathbb R$ then $x^2 = $a negative number is impossible)

If $k=\sqrt{a}\sqrt{b}>0$ (presuming $a,b> 0$) then $k^2 = (\sqrt{a}\sqrt{b})^2 = \sqrt{a}^2 \sqrt{b}^2 = ab$ so if we ask "What is $\sqrt {ab}$, that is what is the unique non-negative number so $x^2=ab$". Well, we just demonstrated that $k = \sqrt{a}\sqrt{b}$ is such a number.

So if we have $M > 0$ and $M$ could be written as $M = a*b$ where $a,b\ge 0$ we would have $\sqrt M = \sqrt{a}\sqrt {b}$.

Now you are correct that $M = (-a)*(-b)$. But how does that help us in anyway. We explained why $M =a*b$ works but $M=(-a)*(-b)$ is as useful for find square roots as noting that $M = j+k$ for some $j$ and $k$. It's just not relevant.

We need $\sqrt{a}$ so that $\sqrt{a}^2 =a$ as well as a $\sqrt{b}$ so that $\sqrt{b}^2 = b$ for this to work and if we tried it with $-a*-b$ we'd need $\sqrt{-a}$ so that $\sqrt{-a}^2 = -a$ and there's just no such thing.

Unless we use complex numbers...

Complex number: There exist $i$ and $-i = (-1)*i$ so that $i^2=-1$. (And $(-i)^2 = i^2 = -1$ as well.

So if $k^2 = 12$ we could do $(\sqrt{3}2)^2 = 3*4 = 12$ so $k=\sqrt 3\sqrt 4=2$ is an acceptable answer. So is $(\sqrt{3}i*2i)^2 = (\sqrt\3i)^2*(2i)^2 = (-3)*(-4) = 12$ but $\sqrt{3}i*2i$ is just another way of writing $\sqrt{3}i*2i = (\sqrt{3}2)*(i^2)= -2\sqrt 3$.

So the two values of $k$ are $2\sqrt 3$ and $-2\sqrt 3$ just as we wanted.

But we can do $\sqrt{12} = \sqrt{-3}\sqrt{-4} = \sqrt 3i* 2i=-2\sqrt 3$.

(I suppose I should point out we tend not to worry about primary square roots in complex numbers because it just opens a can of worms.)

  1. "I've been told to take the square root of both sides in the following equation"

Well, short answer, since we do not know whether $x+8$ is positive or negative we don't know that $\sqrt{(x+8)^2} = x+8$. If $x = -17$ that is false.

$\sqrt{(-17+8)^2} = \sqrt{(-9)^2} =\sqrt{81}= 9\ne -17+8$.

Instead all we know is that $\sqrt{(x+8^2} = |x+8|$

$\sqrt{(-17+8)^2}=\sqrt{81} = 9 $ which does $|-9| = |-17+8|$.

So to solve

$(x + 8)^2 = 1 \implies \sqrt{(x+8)^2} = \sqrt 1\implies |x+8| = 1$

$\implies -(x+8) = 1$ OR $x+8 = 1$ $\implies$

$x+8 = -1$ ** OR ** $x+8 = 1$

And we can write that last line as

$x+8 = \pm 1$.

Or we can take a short cut and go directly to:

$(x+8)^2 = 1$

$\sqrt{(8+x)^2} = \sqrt 1$

$8+ x = \pm 1$.

.... and from here we can go

$x = \pm 1 -8$ so $x= -1 -8=-9$ OR $x=1-8=-7$.

$\endgroup$
1
  • $\begingroup$ Dammit!!!!! We do these 5 year old threads KEEP showing up as active an pertinent on the main lists????? Yeah, I know someone editted it an hour ago (Why?) But sheesh the pertinence of that should really not be valued as highly as a new thread. $\endgroup$ – fleablood Jun 14 '20 at 3:21
-1
$\begingroup$

Consider Use of conventional exponentiation symbol ^(1/2) .

For example $1^\frac{1}{2}$ is multivalue for both -1 and 1.

And $(-1)^\frac{1}{2}$ for -i and i.

It may be used to build multivalue expressions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.