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Section 2-12 : Polynomial Inequalities

5. Solve the following inequality.

\[{y^2} - 2y + 1 \le 0\]

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Start Solution

The first thing we need to do is get a zero on one side of the inequality (which is already done for this problem) and then, if possible, factor the polynomial.

\[{\left( {y - 1} \right)^2} \le 0\]
Hint : Where are the only places where the polynomial might change signs?
Show Step 2

Despite the fact that this is an inequality we first need to know where the polynomial is zero. From the factored from we can quickly see that the polynomial will be zero at,

\[y = 1\]
Hint : Is it possible for the polynomial to ever be negative?
Show Step 3

This problem works a little differently than the others in this section. Because the polynomial is a perfect square we know that it can never be negative! It is only possible for it to be zero or positive.

We are being asked to determine where the polynomial is negative or zero. As noted however it isn’t possible for it to be negative. Therefore the only solution we can get for this inequality is where it is zero and we found that in the previous step.

The answer is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{y = 1}}\]

In this case the answer is a single number and not an inequality. This happens on occasion and we shouldn’t worry about these kinds of “unusual” answers.