How to solve quadratic inequalities with two boundariesWhat is possible values of x if the perimeter of some rectangle tile is between 20 cm and 54 cm.

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Brendon Stein · Accepted Answer

Step 1The values of x that verify the inequality, will verify the following inequalities at the same time:20&amp;lt;2x2+12x, 2x2+12x&amp;lt;54So all we have to do is work with both and we will intersect the solution spaces, getting the solution of the original inequality. Let&#039;s start with the first one:20&amp;lt;2x2+12x⇒0&amp;lt;(x+3+19)⋅(x+3−19)⇒x≻3+19 lor x&amp;lt;−3−19The second one will have the following result:2x2+12x&amp;lt;54⇒x2+6x−27&amp;lt;0⇒−9&amp;lt;x&amp;lt;3The intersection is all of the x∈R such that:−9&amp;lt;x&amp;lt;−3−19 lor−3+19&amp;lt;x&amp;lt;3

Alisha Chambers · Accepted Answer

Step 1If start from20&amp;lt;2x2+12x&amp;lt;54that means you need both of the following to be true:1) 2x2+12x−20&amp;gt;02) 2x2+12x−54&amp;lt;0The first quadratic inequality is true for x&amp;lt;−3−19 or for x&amp;gt;−3+19 The second quadratic inequality is true for −9&amp;lt;x&amp;lt;3. You can get to these solutions by graphing the quadratic or using the quadratic formula, then checking signs, as you mention in your question.Combining everything, we need:(x&amp;lt;−3−19 ∪ x&amp;gt;−3+19) ∩ (−9&amp;lt;x&amp;lt;3),where ∪ means &quot;or&quot; and ∩ means &quot;and.&quot;Then, −3−19≈−7.4 and −3+19≈1.4. So, we end up at the following set of values:−9&amp;lt;x&amp;lt;−3−19 ∪ −3+19&amp;lt;x&amp;lt;3.If that last step doesn&#039;t make sense and it&#039;s hard to visualize, try plotting both of the inequalities on a number line, and wherever they&#039;re both true, that&#039;s the solution.

How to solve quadratic inequalities with two boundaries What

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