Algebra - Solutions and Solution Sets

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Section 2.1 : Solutions and Solution Sets

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7. Is \(x = 1\) a solution to \({\left( {x + 1} \right)^2} > 3x + 1\)?

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There really isn’t all that much to do for these kinds of problems. All we need to do is plug the given number into both sides of the inequality and check to see if the inequality is true. In this case that will mean checking to see if the left side is greater than the right side.

Here is that work for this particular problem.

\[\begin{align*}{\left( {1 + 1} \right)^2} & \mathop > \limits^? 3\left( 1 \right) + 1\\ 4\require{cancel} & \cancel{ > }4\,\,\,\,{\mbox{NOT OK}}\end{align*}\]

Be very careful with this type of problem! Four is not greater than 4 (it’s equal to 4 – big difference here) and so, we can see that \(x = 1\) is not a solution to this inequality.

Contrast the inequality in this problem with,

\[{\left( {x + 1} \right)^2} \ge 3x + 1\]

While \(x = 1\) is not a solution to the inequality in the problem statement it is a solution to this inequality since 4 is in fact greater than or equal to 4. The presence of the equal sign in the inequality can make all the difference in the world and we really need to be on the lookout for it. It is easy to miss when it’s there and it is easy to sometimes assume it is there when in fact it isn’t.