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e.g. the equation $|x-2|=3$, geometrically it's easy to solve drawing the number line. I get the solution set {-1,5} which I understand.

but I was looking at some notes I found online of how to solve this algebraically but I don't understand the logic of this method.

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OK, I see that they have put the condition $if$ $x-2 \geq 0$ but why ? and also they multiplied $x-2$ by $-1$ and then come out with the condition $if$ $ x-2<0.$ Why ?

After that if $ x \geq 2$ then $3 = x-2$ which gives us the result of $5$ and the same thing happens if $x<2$ the $3=x-2$ which in turns gives -1.

I'm just trying understand the logic of this method in relations to absolute value and distance, any help would be greatly appreciated.

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    $\begingroup$ "But why?": because $\lvert a\rvert=\begin{cases}a&\text{if }a\ge0\\ -a&\text{if }a<0\end{cases}$. $\endgroup$
    – user239203
    Commented Mar 20, 2020 at 20:55
  • $\begingroup$ How would you explain to a child that $|7| =7$ and $|-7| = 7$. And if $a = -7$ then $|a| = ....$ what. Does it equal $a = -7$? No it equals $7$ where is .. what.... compared to $a$. (Hint: $-(-7) = 7$.) $\endgroup$
    – fleablood
    Commented Mar 21, 2020 at 0:08

4 Answers 4

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If $a\ge 0$ then what is $|a|$? It is $a$.

And if $a < 0$ then what is $|a|$? It is "the positive absolute value that is the size of $a$" (more or less). But what is that in terms of $a$? As $a =(1)\cdot a $ is negative but $-a = (-1)\cdot a$ is ... positive. So $|a| = -a$ if $a < 0$.

Because.... if $a < 0$ then $0 < -a$ and $-a$ is the positive value so

Definition: The absolute value of $a$ which is written as $|a|$ is a non-negative real number so that $|a|=\begin{cases}a& \text{if }a \ge 0\\-a&\text{if }a < 0\end{cases}$.

(In either case $|a| \ge 0$.)

With that in mind the logic is obvious.

...

$|x-2| =3$.

There are two possibilities: either $x-2 \ge 0$ or .... it isn't .....

If $x-2 \ge 0$ then $|x-2| =x-2$ and $|x-2| =3$ means $x-2 = 3$ which means $x=5$.

And if $x-2$ is not greater of equal to $0$ then $x-2 < 0$ and if $x-2 < 0$, then $|x-2| = -(x-2) = 2-x$. And therefore $|x-2|=3$ means $|x-2| = 2-x =3$ and so $x=-1$.

It's that simple and obvious.

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Different approaches possible. One approach is more of a basic geometric approach: To define the absolute value of a number as the distance from that number to $0$ on the number line. So $|8|=8$ because $8$ is eight units away from $0$. And for exactly that reason we also have $|-8|=8$, because minus eight is also $8$ units away from zero. With this interpretation, we can solve your absolute value equation, if we for a moment consider $|z|=3$. From the definition it immediately follows that we have $z=3$ and $z=-3$. Now with $z=x-2$ you find the values of $x$

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  • $\begingroup$ the op said s/he understood it geometrically. S/he wanted the algebraic method explained. $\endgroup$
    – fleablood
    Commented Mar 21, 2020 at 0:19
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Let

$f\left(x\right)=\left|x-2\right|$,

which is to say:

$f\left(x\right)=\sqrt{\left(x-2\right)^{2}}$.

Because $u^{2}\ge0$ for any real $u$, and $\sqrt{u}\ge0$ everywhere it's defined:

$\left|u\right|\ge0$

whether $u$ is positive or negative. Therefore, if

$x-2\ge0$,

the squaring and square root operations simply cancel to give:

$f(x)=x-2$.

However, if

$x-2<0$,

it must be made positive, necessitating:

$f(x)=-(x-2)=2-x$.

The reason they change it to a piecewise is because the resulting functions are much easier to work with and visualize in many cases.

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Set : number $A = x-2$.

Saying that $|A| = 3$ means that the distance of number $A$ from $0$ is $3$.

This implies that : either number $A = +3$ or number $A = -3$. ( Only $+3$ and $-3$ are at a $3$-unit distance from $0$).

Hence two cases ( substituting $x-2$ for A) : (1) $x-2 = 3$ OR (2) $x-2 = -3$.

( More algebraically : $(|A| = 3) \equiv (A=3 \lor -A = 3) \equiv (A=3 \lor A=-3)$,

with symbol $\equiv$ meaning " is equivalent to" and symbol $\lor$ meaning " or")

In case (1) $x-2 +(2) = x = 3+(2) = 5$.

In case (2) $x-2 +(2) =x = -3 +(2) = -1$.

So, the solution set is $S =$ {$ -1, 5$}

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