Question

Please explain with working!!! Find the set of values of x that satisfy the inequality 9x^2-15x<6

Answer 1

`\displaystyle -(1)/(3) < x < 2`

Explanation:

We want to solve the quadratic inequality:

`9x^2-15x<6`

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:

`9x^2-15x=6`

Solve for x:

`\displaystyle \begin{aligned} 9x^2-15x-6 &= 0\\ 3x^2 - 15 -6 &= 0 \\ (3x+1)(x-2) &= 0\end{aligned}`

By the Zero Product Property:

`\displaystyle 3x + 1 = 0 \text{ or } x - 2 = 0`

Hence:

`\displaystyle x = -(1)/(3)\text{ or } x = 2`

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:

`\displaystyle \begin{aligned}9(-1)^2 -15(-1) \, &?\, 6 \\ 9+15 \, &? \, 6 \\ 24&> 6 \xmark \end{aligned}`

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:

`\displaystyle \begin{aligned} 9(0)^2 - 15(0) \, &? \, 6 \\ (0) - (0) \, &? \, 6 \\ 0 &< 6\, \checkmark\end{aligned}`

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:

`\displaystyle \begin{aligned} 9(3)^2 - 15(3) \, &? \, 6 \\ (81) - (45) \, &? \, 6 \\ 36 &> 6\end{aligned}`

Again, this is not our desired symbol.

In conclusion, our only solution is the interval:

`\displaystyle \left(-(1)/(3), 2\right)`

Or as an inequality:

`\displaystyle -(1)/(3) < x < 2`