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13 Mar 2016, 05:58
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Example (x+3)(x−2)(x+3)(x−2) ≤ 2
(x+3)(x−2)(x+3)(x−2) ≤ 2
(x+3)(x−2)(x+3)(x−2) - 2 ≤ 0
(x+3−2x+4)(x−2)(x+3−2x+4)(x−2) ≤ 0
(−x+7)(x−2)(−x+7)(x−2) ≤ 0
(x−7)(x−2)(x−7)(x−2) ≥ 0 -------------------[Multiplying both sides by -1]
x-7 ≥ 0 ---------> x ≥ 7
x-2 ≥ 0 ---------> x ≥ 2
x = 7 and 2 are the critical points.
Zoom in (real dimensions: 912 x 131)Image
The Real line is divided into three regions. Since the Inequation (obtained in step 4) possesses greater than sign which means that LHS of the Inequation is positive. So, the solution set of the given Inequation is the union of the regions containing positive sign.
Hence x ≥ 7 or x ≤ 2 -----------> (-∞, 2] U [7, ∞)
Please assist as I believe equal to two should not be included in the answer set
(x+3)(x−2)(x+3)(x−2) ≤ 2
(x+3)(x−2)(x+3)(x−2) - 2 ≤ 0
(x+3−2x+4)(x−2)(x+3−2x+4)(x−2) ≤ 0
(−x+7)(x−2)(−x+7)(x−2) ≤ 0
(x−7)(x−2)(x−7)(x−2) ≥ 0 -------------------[Multiplying both sides by -1]
x-7 ≥ 0 ---------> x ≥ 7
x-2 ≥ 0 ---------> x ≥ 2
x = 7 and 2 are the critical points.
Zoom in (real dimensions: 912 x 131)Image
The Real line is divided into three regions. Since the Inequation (obtained in step 4) possesses greater than sign which means that LHS of the Inequation is positive. So, the solution set of the given Inequation is the union of the regions containing positive sign.
Hence x ≥ 7 or x ≤ 2 -----------> (-∞, 2] U [7, ∞)
Please assist as I believe equal to two should not be included in the answer set
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ENGRTOMBA2018
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13 Mar 2016, 06:11
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RohitPrakash88
Your solution is NOT correct (the text in red can NOT be obtained when you manipulate the given equation). Look below.
(x+3)(x−2)(x+3)(x−2) ≤ 2 ---> (x+3)(x−2)(x+3)(x−2) - 2 ≤ 0 ---> \([(x+3)(x−2)]^2 - (\sqrt{2})^2 ≤ 0\)
Now apply the formula, \(a^2-b^2 = (a+b)(a-b)\)
Thus, \([(x+3)(x−2)]^2 - (\sqrt{2})^2 ≤ 0\) ---> \([(x+3)(x-2)+\sqrt{2}][(x+3)(x-2)-\sqrt{2}] ≤ 0\), without solving it, you can see that all the 4 roots \(a_1\), \(a_2\),
\(a_3\), \(a_4\) will be of the form \((x-a_1)*(x-a_2)*(x-a_3)*(x-a_4)\) such \(a_1 < a_2 < a_3 < a_4\)
Thus, \([(x+3)(x-2)+\sqrt{2}][(x+3)(x-2)-\sqrt{2}] ≤ 0\) ----> \((x-a_1)*(x-a_2)*(x-a_3)*(x-a_4) ≤ 0\), for this inequality to hold,
\(a_1 ≤ x ≤ a_2\) and \(a_3 ≤ x ≤ a_4\)
Hope this helps.
What is the source of this question?
26 Mar 2016, 14:54
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RohitPrakash88
Engr2012 has a more detailed solution, but I'll add this. Always check your answer to a problem like this before committing to it. In this case, it's easy to decide whether x = 2 belongs in the set of solutions. Just plug 2 into the original equation -- if it simplifies to a true statement, then 2 is a valid solution. In this case, it is.
(2+3)(2−2)(2+3)(2−2) ≤ 2 ?
(5)(0)(5)(0) ≤ 2?
0 ≤ 2?
So, x = 2 is a valid solution to the inequality. In fact, if something like this appeared on the GMAT, it would almost certainly be amenable to plugging in and checking the answer choices, rather than trying to simplify algebraically.
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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