# Absurd result in inequality

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?

• $x-2$ can be negative. – razivo Nov 28 '20 at 16:22
• 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. – JMoravitz Nov 28 '20 at 16:22
• "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 $(x-2)^2\geq 0$ – JMoravitz Nov 28 '20 at 16:23
• Thank you I got it! – E. Williams Nov 28 '20 at 16:23
• 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. – Peter Szilas Nov 28 '20 at 16:27

You couldn’t have divided both sides by $$x-2$$ As it can be negative. $$-1>-2 \not \Rightarrow 1>2$$
• "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. – JMoravitz Nov 28 '20 at 16:25
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.