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28 Nov 2023, 09:08
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TargetMBA007
If I understand your doubt correctly, the problem is that in order for (x+1)(x−1) ≤ 0 to be true, (x+1) and (x−1) must have opposite signs, as discussed in the third solution above. Hence, (x+1) ≤ 0 and (x−1) ≤ 0 does not make sense: if both are negative, which happens when x ≤ -1, their product will be positive, not negative.
Does this answer your question?
TargetMBA007
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28 Nov 2023, 18:25
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Bunuel - Thanks Bunuel, I think I am finally getting there. I would just like to clarify my approach below.
My guess is, in each case (more than/less than) we would simply set each of the factors against opposing signs?
SMALLER THAN SCENARIO
With this (x+1)(x−1) ≤ 0 its easy, as the ans would be -1 ≤ x ≤ 1, but what if the factors were different, say
(X+3)(x-2) ≤ 0, how do we decide which factor is set against less than sign, and which factor against more than sign to find the final answer - I am guessing it is always the smaller number that is set against the < and the bigger number against the > sign.
So here we will have:
(x-2) ≤ 0
(X+3)≥ 0
Hence,
-3≤x≤ 2, is that right?
GREATER THAN SCENARIO
for > sign. I believe it would be opposite, and we set the smaller number against the > sign, and bigger number against the < than sign? say
(x-2) ≥ 0
(X+3)≤ 0
Hence X≤-3 or x≥2?
Hopefully I am making sense?
My guess is, in each case (more than/less than) we would simply set each of the factors against opposing signs?
SMALLER THAN SCENARIO
With this (x+1)(x−1) ≤ 0 its easy, as the ans would be -1 ≤ x ≤ 1, but what if the factors were different, say
(X+3)(x-2) ≤ 0, how do we decide which factor is set against less than sign, and which factor against more than sign to find the final answer - I am guessing it is always the smaller number that is set against the < and the bigger number against the > sign.
So here we will have:
(x-2) ≤ 0
(X+3)≥ 0
Hence,
-3≤x≤ 2, is that right?
GREATER THAN SCENARIO
for > sign. I believe it would be opposite, and we set the smaller number against the > sign, and bigger number against the < than sign? say
(x-2) ≥ 0
(X+3)≤ 0
Hence X≤-3 or x≥2?
Hopefully I am making sense?
29 Nov 2023, 00:22
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TargetMBA007
EXAMPLE 1:
For (x+3)(x-2) ≤ 0 to be true, (x+3) and (x-2) must have opposite signs.
- 1. x + 3 ≥ 0 and x - 2 ≤ 0 give x ≥ -3 and x ≤ 2, which leads to -3 ≤ x ≤ 2.
2. x + 3 ≤ 0 and x - 2 ≥ 0 is not possible since if x + 3, the larger factor, is negative, then x - 2, the smaller factor, will also be negative.
So, (x+3)(x-2) ≤ 0 means that -3 ≤ x ≤ 2.
EXAMPLE 2:
For (x+3)(x-2) ≥ 0 to be true, (x+3) and (x-2) must have the same sign.
- 1. x + 3 ≥ 0 and x - 2 ≥ 0 gives x ≥ -3 and x ≥ 2, which means x ≥ 2.
2. x + 3 ≤ 0 and x - 2 ≤ 0 gives x ≤ -3 and x ≤ 2, which means x ≤ -3.
So, (x+3)(x-2) ≥ 0 means that x ≤ -3 or x ≥ 2.
If you are not sure about this approach, then I would advise studying the Wavy Line Method for solving inequalities.
(x+3)(x-2) < 0
Here is how to solve the above inequality easily. The "roots", in ascending order, are -3, and 2, which gives us 3 ranges:
- \(x < -3\);
\(-3 < x < 2\);
\(x > 2\).
Next, test an extreme value for \(x\): if \(x\) is some large enough number, say 10, then both multiples will be positive, giving a positive result for the whole expression. So when \(x > 2\), the expression is positive. Now the trick: as in the 3rd range, the expression is positive, then in the 2nd it'll be negative, and finally in the 1st, it'll be positive: \(\text{(+ - +)}\). So, the ranges when the expression is negative is: \(-3 < x < 2\).
P.S. You can apply this technique to any inequality of the form \((ax - b)(cx - d)(ex - f)... > 0\) or \( < 0\). If one of the factors is in the form \((b - ax)\) instead of \((ax - b)\), simply rewrite it as \(-(ax - b)\) and adjust the inequality sign accordingly. For example, consider the inequality \((4 - x)(2 + x) > 0\). Rewrite it as \(-(x - 4)(2 + x) > 0\). Multiply by -1 and flip the sign: \((x - 4)(2 + x) < 0\). The "roots", in ascending order, are -2 and 4, giving us three ranges: \(x < -2\), \(-2 < x < 4\), and \(x > 4\). If \(x\) is some large enough number, say 100, both factors will be positive, yielding a positive result for the entire expression. Thus, when \(x > 4\), the expression is positive. Now, apply the alternating sign trick: as the expression is positive in the 3rd range, it will be negative in the 2nd range, and positive in the 1st range: \(\text{(+ - +)}\). Therefore, the ranges where the expression is negative is: \(-2 < x < 4\).
Check the links below:
9. Inequalities
- Theory
Inequalities Made Easy!
Inequalities: Tips and hints
Arithmetic with Inequalities
Wavy Line Method Application - Complex Algebraic Inequalities
Solving Quadratic Inequalities: Graphic Approach
Inequalities - Quadratic Inequalities
Graphic approach to problems with inequalities
Inequalities trick
INEQUATIONS(Inequalities) Part 1
INEQUATIONS(Inequalities) Part 2
Hope it helps.
