Solving the Inequality: How to Find the Solution for (a-x+1)(a-x+2) ≤ a?

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SUMMARY

The inequality (a-x+1)(a-x+2) ≤ a can be solved by expanding the left-hand side and rearranging it entirely in terms of x. Key steps include recognizing that the left side can be expressed as a difference of two squares. To isolate x in terms of a, the expression can be rewritten as ((a - x + 3/2) - 1/2)((a - x + 3/2) + 1/2) ≤ a. Understanding the rules of inequalities, such as not dividing by zero and flipping the inequality sign when dividing by a negative number, is crucial for accurate manipulation.

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Students, educators, and anyone interested in mastering algebraic inequalities and polynomial manipulation will benefit from this discussion.

nightking
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How can I solve this inequality?

(a-x+1)(a-x+2) ≤ a

where a is a constant with unknown value.

Thanks in advance.
 
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Hey nightking and welcome to the forums.

You need to expand out the left hand side and then put on side completely in terms of x.

The rules for inequalities are that you can't divide any side by zero (you also have to make sure any variables you have are not zero either if you want to divide), if you divide by a negative number you flip the inequality sign, if you subtract or add a term the sign doesn't change.
 
If you want to find x in terms of a, I would start with
((a - x + 3/2) - 1/2)((a - x + 3/2) + 1/2) ≤ a

The left hand side is then the difference of two squares...
 
AlephZero said:
If you want to find x in terms of a, I would start with
((a - x + 3/2) - 1/2)((a - x + 3/2) + 1/2) ≤ a

The left hand side is then the difference of two squares...

Brilliant. Thanks!
 

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