Range of x for which the modulus inequality will be valid: |x2−2x−8|&gt;2x: Need clarification about solution set

Question

Range of x for which the modulus inequality will be valid: |x2−2x−8|&amp;gt;2x: Need clarification about solution set

Marcos Boyer · Accepted Answer

You seem to have taken the range common to the two solution sets. In other words, you have taken the intersection of the two sets, which is also done incorrectly because the intersection of the two cases you have taken is a null set because those two conditions cannot be true simultaneously. x2−2x−8≥0 and x2−2x−8&amp;lt;0 is not possible.{x2−2x−8≥0}∼∩∼{x2−2x−8&amp;lt;0}=ϕThe final solution set that you are looking for is the union of the two sets because either x2−2x−8≥0 or x2−2x−8&amp;lt;0. Or signifies that you are supposed to take the union of the sets.{x2−2x−8≥0}∼∪∼{x2−2x−8&amp;lt;0}=(−∞,22)∪(2+23,∞).

Kingston Lowery · Accepted Answer

Consider the first part. You wrote that you were dealing with the case in which&amp;nbsp;x2−2x−8≥0. That means that&amp;nbsp;x≤−2&amp;nbsp;or that&amp;nbsp;x≥4. Then you solved the inequation&amp;nbsp;x2−2x+8&amp;gt;2x, and you should have got that&amp;nbsp;x&amp;gt;2+23&amp;nbsp;or that&amp;nbsp;x&amp;lt;2−23. So, the solution when you are in the first situation is that&amp;nbsp;x∈(−∞,−2]∪[2+23,∞).Now, the second part. Asserting that&amp;nbsp;x2−2x−8&amp;lt;0&amp;nbsp;means that&amp;nbsp;x∈(−2,4). And asserting that&amp;nbsp;−(x2−2x−8)&amp;gt;2x&amp;nbsp;means that&amp;nbsp;x∈(−22,22). So, the solution when you are in the second situation is that&amp;nbsp;x∈(−2,22).So, the global answer is that&amp;nbsp;x∈(−∞,22]∪[4,∞).

Range of x for which the modulus inequality

Answered question

Answer & Explanation

New Questions in Algebra I