Answer :
To solve the polynomial inequality [tex]\(5x^2 + 13x - 6 < 0\)[/tex], let's work through the problem step by step.
### Step 1: Rewrite the Inequality
First, we start with the given inequality:
[tex]\[ 5x^2 + 13x - 6 < 0 \][/tex]
### Step 2: Factor the Polynomial
Next, we need to find the roots of the polynomial [tex]\( 5x^2 + 13x - 6 = 0 \)[/tex] because these roots will help us determine the intervals for testing the inequality.
The roots are found by solving for [tex]\( x \)[/tex] in the equation:
[tex]\[ 5x^2 + 13x - 6 = 0 \][/tex]
Let's denote the roots by [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]. We can use the quadratic formula to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 5 \)[/tex], [tex]\( b = 13 \)[/tex], and [tex]\( c = -6 \)[/tex].
However, solving this quadratic equation exactly will yield irrational or fractional roots, so we'll use these roots to determine the nature of the solution.
### Step 3: Determine the Intervals
Once we have the roots, let's denote them as [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], where [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] are the solutions to the quadratic equation. The inequality [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex] holds true within the intervals formed by these roots.
Given the roots [tex]\( x_1 = -3 \)[/tex] and [tex]\( x_2 = \frac{2}{5} \)[/tex], we need to check the sign of the polynomial in the intervals determined by [tex]\( (-\infty, x_1) \)[/tex], [tex]\( (x_1, x_2) \)[/tex], and [tex]\( (x_2, \infty) \)[/tex].
### Step 4: Test the Intervals
We need to test the function [tex]\( 5x^2 + 13x - 6 \)[/tex] within the intervals formed by the roots.
For [tex]\( (-\infty, -3) \)[/tex]:
- Test point [tex]\( x = -4 \)[/tex]
[tex]\[ 5(-4)^2 + 13(-4) - 6 = 80 - 52 - 6 = 22 > 0 \][/tex]
This interval does not satisfy [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
For [tex]\( (-3, \frac{2}{5}) \)[/tex]:
- Test point [tex]\( x = 0 \)[/tex]
[tex]\[ 5(0)^2 + 13(0) - 6 = -6 < 0 \][/tex]
This interval satisfies [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
For [tex]\( (\frac{2}{5}, \infty) \)[/tex]:
- Test point [tex]\( x = 1 \)[/tex]
[tex]\[ 5(1)^2 + 13(1) - 6 = 5 + 13 - 6 = 12 > 0 \][/tex]
This interval does not satisfy [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
### Step 5: Express the Solution in Interval Notation
The polynomial inequality is satisfied in the interval [tex]\( (-3, \frac{2}{5}) \)[/tex].
### Solution Set
Thus, the solution set in interval notation is:
[tex]\[ \boxed{(-3, \frac{2}{5})} \][/tex]
### Graph the Solution Set
To graph the solution set on a real number line, we put open circles on [tex]\( -3 \)[/tex] and [tex]\( \frac{2}{5} \)[/tex] as these endpoints are not included in the solution set, and shade the region between them.
This completes the solution to the inequality [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
### Step 1: Rewrite the Inequality
First, we start with the given inequality:
[tex]\[ 5x^2 + 13x - 6 < 0 \][/tex]
### Step 2: Factor the Polynomial
Next, we need to find the roots of the polynomial [tex]\( 5x^2 + 13x - 6 = 0 \)[/tex] because these roots will help us determine the intervals for testing the inequality.
The roots are found by solving for [tex]\( x \)[/tex] in the equation:
[tex]\[ 5x^2 + 13x - 6 = 0 \][/tex]
Let's denote the roots by [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]. We can use the quadratic formula to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 5 \)[/tex], [tex]\( b = 13 \)[/tex], and [tex]\( c = -6 \)[/tex].
However, solving this quadratic equation exactly will yield irrational or fractional roots, so we'll use these roots to determine the nature of the solution.
### Step 3: Determine the Intervals
Once we have the roots, let's denote them as [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], where [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] are the solutions to the quadratic equation. The inequality [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex] holds true within the intervals formed by these roots.
Given the roots [tex]\( x_1 = -3 \)[/tex] and [tex]\( x_2 = \frac{2}{5} \)[/tex], we need to check the sign of the polynomial in the intervals determined by [tex]\( (-\infty, x_1) \)[/tex], [tex]\( (x_1, x_2) \)[/tex], and [tex]\( (x_2, \infty) \)[/tex].
### Step 4: Test the Intervals
We need to test the function [tex]\( 5x^2 + 13x - 6 \)[/tex] within the intervals formed by the roots.
For [tex]\( (-\infty, -3) \)[/tex]:
- Test point [tex]\( x = -4 \)[/tex]
[tex]\[ 5(-4)^2 + 13(-4) - 6 = 80 - 52 - 6 = 22 > 0 \][/tex]
This interval does not satisfy [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
For [tex]\( (-3, \frac{2}{5}) \)[/tex]:
- Test point [tex]\( x = 0 \)[/tex]
[tex]\[ 5(0)^2 + 13(0) - 6 = -6 < 0 \][/tex]
This interval satisfies [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
For [tex]\( (\frac{2}{5}, \infty) \)[/tex]:
- Test point [tex]\( x = 1 \)[/tex]
[tex]\[ 5(1)^2 + 13(1) - 6 = 5 + 13 - 6 = 12 > 0 \][/tex]
This interval does not satisfy [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
### Step 5: Express the Solution in Interval Notation
The polynomial inequality is satisfied in the interval [tex]\( (-3, \frac{2}{5}) \)[/tex].
### Solution Set
Thus, the solution set in interval notation is:
[tex]\[ \boxed{(-3, \frac{2}{5})} \][/tex]
### Graph the Solution Set
To graph the solution set on a real number line, we put open circles on [tex]\( -3 \)[/tex] and [tex]\( \frac{2}{5} \)[/tex] as these endpoints are not included in the solution set, and shade the region between them.
This completes the solution to the inequality [tex]\( 5x^2 + 13x - 6 < 0 \)[/tex].
