Answer:
[tex]\displaystyle -\frac{1}{3} < x < 2[/tex]
Step-by-step explanation:
We want to solve the quadratic inequality:
[tex]9x^2-15x<6[/tex]
In order to solve a quadratic inequality, find its zeros. That is, let the inequality sign be an equal sign and solve for x. We do this because it denotes when the graph crosses the x-axis: that is, when it is positive and/or negative. This yields:
[tex]9x^2-15x=6[/tex]
Solve for x:
[tex]\displaystyle \begin{aligned} 9x^2-15x-6 &= 0\\ 3x^2 - 15 -6 &= 0 \\ (3x+1)(x-2) &= 0\end{aligned}[/tex]
By the Zero Product Property:
[tex]\displaystyle 3x + 1 = 0 \text{ or } x - 2 = 0[/tex]
Hence:
[tex]\displaystyle x = -\frac{1}{3}\text{ or } x = 2[/tex]
Now, we can test intervals. The three intervals are: all values less than -1/3, values between -1/3 and 2, and all values greater than 2.
To test the first interval, let x = -1. Substitute this into our original inequality:
[tex]\displaystyle \begin{aligned}9(-1)^2 -15(-1) \, &?\, 6 \\ 9+15 \, &? \, 6 \\ 24&> 6 \xmark \end{aligned}[/tex]
The resulting symbol is "greater than," which is not our desired symbol.
To test the interval between -1/3 and 2, we can let x = 0:
[tex]\displaystyle \begin{aligned} 9(0)^2 - 15(0) \, &? \, 6 \\ (0) - (0) \, &? \, 6 \\ 0 &< 6\, \checkmark\end{aligned}[/tex]
The resulting symbol is indeed less than. So, the interval (-1/3, 2) is a part of our solution.
Finally, to test the third interval, let x = 3. Then:
[tex]\displaystyle \begin{aligned} 9(3)^2 - 15(3) \, &? \, 6 \\ (81) - (45) \, &? \, 6 \\ 36 &> 6\end{aligned}[/tex]
Again, this is not our desired symbol.
In conclusion, our only solution is the interval:
[tex]\displaystyle \left(-\frac{1}{3}, 2\right)[/tex]
Or as an inequality:
[tex]\displaystyle -\frac{1}{3} < x < 2[/tex]
