0
$\begingroup$

I have this inequality

$(x-2)^2 \ge 0 $

So I solve it like this

$(x-2)^2 \ge 0 \implies (x-2)(x-2)\ge0$

$\implies x(x-2)-2(x-2)\ge0 \implies x(x-2)\ge2(x-2)\implies x\ge2$

But obviously $x\ge2$ is false in R

What did I miss?

$\endgroup$
5
  • 2
    $\begingroup$ $x-2$ can be negative. $\endgroup$ – razivo Nov 28 '20 at 16:22
  • 3
    $\begingroup$ You divided both sides by $x-2$... What happens if $x-2$ was equal to zero? You just divided both sides by zero. You can't do that. What if $x-2$ is positive? Well... that is fine, you had the right outcome. What if $x-2$ was negative? Well... the sign should have flipped. $\endgroup$ – JMoravitz Nov 28 '20 at 16:22
  • 1
    $\begingroup$ "I have this inequality: $(x-2)^2\geq 0$" You should know that a real number squared is always greater than or equal to zero... so every $x$ satisfies $(x-2)^2\geq 0$ $\endgroup$ – JMoravitz Nov 28 '20 at 16:23
  • $\begingroup$ Thank you I got it! $\endgroup$ – E. Williams Nov 28 '20 at 16:23
  • $\begingroup$ You divide by (x-2); If (x-2)>0, then you get x>2. Division by (x-2)<0 (negative, the > changes into <) you get x<2. $\endgroup$ – Peter Szilas Nov 28 '20 at 16:27
0
$\begingroup$

You couldn’t have divided both sides by $x-2$ As it can be negative. $$-1>-2 \not \Rightarrow 1>2$$

$\endgroup$
2
  • 2
    $\begingroup$ "You couldn't have divided..." Yes, you could... but in doing so you need to split into cases based on the sign of $x-2$ and considered each case separately. $\endgroup$ – JMoravitz Nov 28 '20 at 16:25
  • $\begingroup$ Yes, maybe “you couldn’t have just have divided” would have fitted better. $\endgroup$ – razivo Nov 28 '20 at 16:32
0
$\begingroup$

In your last step you divide both sides of the inequality by $(x-2)$ which is negative for $x < 2$, and this would require reversing the inequality sign. Instead you could simply split your solution into two cases: $x\geq 2$ and $x< 2$, and derive the entire solution space by considering each case individually.

$\endgroup$

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.