Doubt solving a polynomial inequality

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SUMMARY

The discussion focuses on solving the polynomial inequality involving the square root function, specifically the inequality $$\sqrt{x^{2}-2x}>x-1$$. The user initially misinterprets the inequality after squaring both sides, leading to incorrect conclusions. Key insights include the importance of considering the sign of both sides before squaring and recognizing that squaring can alter the direction of the inequality. The correct approach involves analyzing the conditions for positivity and applying the union of intervals where the inequality holds true.

PREREQUISITES
  • Understanding of polynomial inequalities
  • Knowledge of square root properties
  • Familiarity with absolute value concepts
  • Basic graphing skills for function analysis
NEXT STEPS
  • Study the rules for manipulating inequalities, especially when squaring both sides
  • Learn about the properties of absolute values in inequalities
  • Explore interval notation and union of intervals in solution sets
  • Practice solving various polynomial inequalities with different conditions
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Mathematics students, educators, and anyone interested in mastering polynomial inequalities and their properties.

greg_rack
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Homework Statement
$$\sqrt{x^{2}-2x}-x+1>0$$
Relevant Equations
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I got this function in a function analysis and got confused on how to solve its positivity;
I rewrote it as:
$$\sqrt{x^{2}-2x}>x-1 \rightarrow x^2-2x>x^2-2x+1$$
And therefore concluded it must've been impossible... but I'm certainly missing something stupid, since plotting the graphs of the two functions(##\sqrt{x^{2}-2x}## and ##x-1##) I see that the first is greater only for ##x<0##.

Maybe the flaw comes when I'm squaring both factors in the inequality... should I put the first factor in an absolute value, since it must be positive, as a square root?
But then, how do I solve it with the abs.?
I can deduce that the left term must be greater until the second term is negative... and for positive values it's the reversed situation, and this thinking works, but is there a more "rigid" procedure?
 
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You need to study the various part this inequality have.

First, x<0 or x>2 is a condition to the root.

So in the case x<0 the inequality is immediate, since the root is always greater than zero, and x-1 is zero

Now you apply for x>2 too.

Take the union.
 
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Your mistake is in the squaring, squaring inequalities just doesn't work sometimes. For example take the inequality -2<1 which is true. If you square it however you get the inequality 4<1 which is false..
 
Delta2 said:
Your mistake is in the squaring, squaring inequalities just doesn't work sometimes. For example take the inequality -2<1 which is true. If you square it however you get the inequality 4<1 which is false..
So, generally, you don't square inequalities? Is there a rule to avoid this type of mistakes?
 
I think the rule about squaring inequalities is that:

if both sides are positive then you can square it and the inequality keeps its direction

if both sides are negative then the inequality reverses direction after the squaring,

and if one side is negative and one positive then squaring is unpredictable!
 
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