Sure, let's break this down step by step to solve both the quadratic equation and the inequality.
### 1. Solve the Quadratic Equation [tex]\( 1.1x^2 - x - 12 = 0 \)[/tex]
The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = 1.1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -12 \)[/tex].
The solutions for [tex]\( x \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's first calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-1)^2 - 4 \cdot 1.1 \cdot (-12) \][/tex]
[tex]\[ \Delta = 1 + 52.8 \][/tex]
[tex]\[ \Delta = 53.8 \][/tex]
Next, we calculate the roots using the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{53.8}}{2 \cdot 1.1} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{53.8}}{2.2} \][/tex]
So, we have:
[tex]\[ x_1 = \frac{1 + \sqrt{53.8}}{2.2} \][/tex]
[tex]\[ x_2 = \frac{1 - \sqrt{53.8}}{2.2} \][/tex]
### 2. Solve the Inequality [tex]\( (2x - 3)(x^2 + 1) < 0 \)[/tex]
To analyze this inequality, let's study the product:
1. [tex]\( x^2 + 1 \)[/tex] is always positive for all real [tex]\( x \)[/tex] because it's a sum of squares.
2. Therefore, [tex]\( (2x - 3)(x^2 + 1) < 0 \)[/tex] will hold true when [tex]\( 2x - 3 < 0 \)[/tex].
Solving the inequality [tex]\( 2x - 3 < 0 \)[/tex]:
[tex]\[ 2x - 3 < 0 \][/tex]
[tex]\[ 2x < 3 \][/tex]
[tex]\[ x < \frac{3}{2} \][/tex]
[tex]\[ x < 1.5 \][/tex]
### Summary of Solutions
- The solutions to the quadratic equation [tex]\( 1.1x^2 - x - 12 = 0 \)[/tex] are:
[tex]\[ x_1 = \frac{1 + \sqrt{53.8}}{2.2}, \quad x_2 = \frac{1 - \sqrt{53.8}}{2.2} \][/tex]
- The solution to the inequality [tex]\( (2x - 3)(x^2 + 1) < 0 \)[/tex] is:
[tex]\[ x < 1.5 \][/tex]
These are the required solutions for the quadratic equation and the inequality.
