Question

(2x+3)(4x-3)(x-10) >0​

Answer 1

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Answer 2

To solve the inequality $(2x+3)(4x-3)(x-10) > 0$, we need to determine the intervals where the expression is greater than zero. We can do this by analyzing the signs of the factors and finding where the expression is positive.

Let's break down the problem step by step:

1. Identify the critical points by setting each factor equal to zero and solving for $x$:

   • $2x + 3 = 0$ gives $x = -\frac{3}{2}$
   • $4x - 3 = 0$ gives $x = \frac{3}{4}$
   • $x - 10 = 0$ gives $x = 10$

2. Now, we have three critical points: $x = -\frac{3}{2}$, $x = \frac{3}{4}$, and $x = 10$. We'll use these points to divide the number line into four intervals:

   • Interval 1: $x < -\frac{3}{2}$
   • Interval 2: $-\frac{3}{2} < x < \frac{3}{4}$
   • Interval 3: $\frac{3}{4} < x < 10$
   • Interval 4: $x > 10$

3. Now, let's analyze the sign of the expression within each interval by testing a point within each interval:

   • For Interval 1 ($x < -\frac{3}{2}$), let's test $x = -2$:

     • $(2(-2)+3)(4(-2)-3)(-2-10) = (-1)(-11)(-12) = 132$, which is positive.

   • For Interval 2 ($-\frac{3}{2} < x < \frac{3}{4}$), let's test $x = 0$:

     • $(2(0)+3)(4(0)-3)(0-10) = (3)(-3)(-10) = 90$, which is positive.

   • For Interval 3 ($\frac{3}{4} < x < 10$), let's test $x = 5$:

     • $(2(5)+3)(4(5)-3)(5-10) = (13)(17)(-5) = -1105$, which is negative.

   • For Interval 4 ($x > 10$), let's test $x = 15$:

     • $(2(15)+3)(4(15)-3)(15-10) = (33)(57)(5) = 9405$, which is positive.

4. Based on the analysis above, we see that the expression is positive in Intervals 1 and 2, and it's negative in Interval 3. In Interval 4, the expression is positive again. Therefore, the solution to the inequality $(2x+3)(4x-3)(x-10) > 0$ is: $x < -\frac{3}{2} \quad \text{or} \quad -\frac{3}{2} < x < \frac{3}{4} \quad \text{or} \quad x > 10.$

Please note that the endpoints of the intervals are not included in the solution because the inequality is strict ($>$).