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?
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2$\begingroup$ $x-2$ can be negative. $\endgroup$ – razivo Nov 28 '20 at 16:22
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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
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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
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$\begingroup$ Thank you I got it! $\endgroup$ – E. Williams Nov 28 '20 at 16:23
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$\begingroup$ 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
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You couldn’t have divided both sides by $x-2$ As it can be negative. $$-1>-2 \not \Rightarrow 1>2$$
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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
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$\begingroup$ Yes, maybe “you couldn’t have just have divided” would have fitted better. $\endgroup$ – razivo Nov 28 '20 at 16:32
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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.
answered Nov 28 '20 at 16:31
