genious2000:
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Any question asking for a common solution requires solving it using wavy curves, a tool, to make stuff easier.
For example,
Equation: (x- \alpha )(x- \beta ) \leq 0(x−α)(x−β)≤0 will have a wavy curve like the first attachment below.
The convention is that the upper loop in a wave of the wavy curve represents where the equation will generate a positive result and vice versa.
Remember, the wavy curve is a number line representation of values, it's NOT a graph of the given equation.
Now, let's draw the wavy curves for the given equations.
(x-7)(x-15)\ \textless \ 0(x−7)(x−15) < 0 will have the wavy curve as attached in the second attachment below.
Let's take a sample value >15, let's take 16.
The given equation with x=16 will yield (16-7)(16-15) = 9 * 1 = 9, positive.
In 7<x<15, let's take x=10. The equation will yield (10-7)(10-15) = 3 * -5 = -15, negative.
For values less than x=7, let's take x=6, the equation is yielding (6-7)(6-15) = -1* -9 = 9, positive.
Clearly, we need to take the <0 part for the required range of the inequality.
According to the first equation, x ∈ (7, 15) ...(i)
The second equation (x-2)(x-8)(x-12) \geq 0(x−2)(x−8)(x−12)≥0 will give a wavy curve as represented by the third attachment.
One can take sample values to check a wavy curve.
Clearly, x∈ [2, 8] ∪ [12, ∞) ...(ii)
The region covered by both (i) and (ii) shall be,
x ∈ (7, 8] ∪ [12, 15)
And that, shall be the answer!
For example,
Equation: (x- \alpha )(x- \beta ) \leq 0(x−α)(x−β)≤0 will have a wavy curve like the first attachment below.
The convention is that the upper loop in a wave of the wavy curve represents where the equation will generate a positive result and vice versa.
Remember, the wavy curve is a number line representation of values, it's NOT a graph of the given equation.
Now, let's draw the wavy curves for the given equations.
(x-7)(x-15)\ \textless \ 0(x−7)(x−15) < 0 will have the wavy curve as attached in the second attachment below.
Let's take a sample value >15, let's take 16.
The given equation with x=16 will yield (16-7)(16-15) = 9 * 1 = 9, positive.
In 7<x<15, let's take x=10. The equation will yield (10-7)(10-15) = 3 * -5 = -15, negative.
For values less than x=7, let's take x=6, the equation is yielding (6-7)(6-15) = -1* -9 = 9, positive.
Clearly, we need to take the <0 part for the required range of the inequality.
According to the first equation, x ∈ (7, 15) ...(i)
The second equation (x-2)(x-8)(x-12) \geq 0(x−2)(x−8)(x−12)≥0 will give a wavy curve as represented by the third attachment.
One can take sample values to check a wavy curve.
Clearly, x∈ [2, 8] ∪ [12, ∞) ...(ii)
The region covered by both (i) and (ii) shall be,
x ∈ (7, 8] ∪ [12, 15)
And that, shall be the answer!
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