-1

Suppose, I have two points $A(x, 6)$ and $B(5,6)$ and $AB = 4$ unit. We have to find the value of $x$.

Process 1:

$$AB=\sqrt{(x-5)^2+0}$$

[Only positive root is taken as length is positive]

$$\implies 4=\sqrt{(x-5)^2}$$

$$\implies 4=x-5$$

$$\implies x=9$$

Although I got one acceptable value of $x$, I couldn't get the other acceptable value of $x$, which is 1.

Process 2:

$$AB=\sqrt{(x-5)^2+0}$$

[Only positive root is taken as length is positive]

$$\implies 4=\sqrt{(x-5)^2}$$

$$\implies 16=(x-5)^2$$

[squaring both sides]

$$\implies x^2-10x+9=0$$

$$\implies x=1,9$$

Why does process 2 work and process 1 doesn't when both of them are legitimate processes?

tryingtobeastoic
  • 3,176

2 Answers2

1

Why does process 2 work and process 1 doesn't when both of them are legitimate processes?

$$\implies 4=\sqrt{(x-5)^2}$$ $$\implies 4=x-5$$

This step is wrong, $\sqrt{a^2}$ does not simplify to $a$, but to $|a|$. Note that if $x-5=-4$, you still have $(x-5)^2=16$ and so $\sqrt{16}=4$. You'll get both solutions continuing from $4=|x-5|$.

StackTD
  • 27,903
  • 34
  • 63
  • Thanks for your response! I took only the positive root in process 1 as AB is always positive. That's why I wrote $$AB=\sqrt{(x-5)^2+0}$$ and not $$AB=\pm\sqrt{(x-5)^2+0}$$ – tryingtobeastoic Aug 11 '21 at 14:10
  • That's fine - the mistake was when simplyfing $\sqrt{(...)^2}$, as I wrote in my answer. – StackTD Aug 11 '21 at 14:22
0

Process 1 is wrong; process 2 is okay for this reason (@ingix's answer is preferable), and thus doesn't produce any extraneous roots. Let me talk about process 3, which is the corrected form of process 1.

Process 3:

$$\sqrt{(x-5)^2+0}=4$$

$$\implies \sqrt{(x-5)^2}=4$$

$$\implies |{x-5}|=4$$

$$\implies x-5=\pm4$$

$$x=1,9$$

Capeesh?

See this (must) and see @HansLundmark's comment in this.

tryingtobeastoic
  • 3,176