inequality reason

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monick

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x(x−1)2⩾(4−x)(x−12⇒x⩾4−xx (x-1)^{2} \geqslant (4-x)(x-1^{2}\Rightarrow x\geqslant 4-xtrue or false
ps can i remove (x−1)2(x-1)^{2} from both sides even if they are zero
 
monick said:
x(x−1)2⩾(4−x)(x−12⇒x⩾4−xx (x-1)^{2} \geqslant (4-x)(x-1^{2}\Rightarrow x\geqslant 4-xtrue or false
ps can i remove (x−1)2(x-1)^{2} from both sides even if they are zero
Click to expand...
No! That would be equivalent to dividing by 0.

What else can you do legally?
 
monick said:
x(x−1)2⩾(4−x)(x−1)2⇒x⩾4−xx (x-1)^{2} \geqslant (4-x)(x-1)^{2}\Rightarrow x\geqslant 4-xtrue or false
ps can i remove (x−1)2(x-1)^{2} from both sides even if they are zero
Click to expand...
Two common ways to avoid illegal or misleading divisions are to replace the division with factoring, or to use separate cases. Either can work here; both will lead to a two-part result.
 
It depends on what you mean by ”remove”?

As Dr. Peterson says, you cannot DIVIDE by (x−1)2(x - 1)^2 if that equals zero.

If you mean REPLACE the expression (x−1)2(x - 1)^2 by zero in order to consider the special case of
(x−1)2=0(x - 1)^2 = 0, you can do that, but be careful.

If you mean EXPAND both the left hand and right hand sides by carrying out the indicated operations, you can do that.

Part of math is being careful in what you mean. “Remove” is not an arithmetic operation.
 
I have a concern with this problem.
Note that x> 4-x <=> x>2.
However x= 1 is a solution, as both sides of the inequality will be zero. My concern is that 1>2 is false.
 
Last edited:
Subhotosh Khan said:
No! That would equivalent to dividing by 0.
Click to expand...
Boss, you are correct that it might be equivalent to dividing by 0.
The way around that is to see if x=1 is a solution or not and then divide both sides by (x-1)^2. If in fact x=1 is a zero, don't forget this in your solution. The author of this problem actually did leave out x=1 as a solution.
 
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