Solve the inequality \( 3 y^{2}+4 y-32>0 \) Show your working clearly.

To solve the inequality \( 3y^{2} + 4y - 32 > 0 \), we'll follow these steps: 1. **Find the Roots of the Equation \( 3y^{2} + 4y - 32 = 0 \)** 2. **Determine the Intervals Defined by the Roots** 3. **Test Each Interval to See Where the Inequality Holds** --- ### 1. Find the Roots of the Equation \( 3y^{2} + 4y - 32 = 0 \) We'll use the **quadratic formula** to find the roots of the quadratic equation: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where: - \( a = 3 \) - \( b = 4 \) - \( c = -32 \) **Calculate the Discriminant (\( D \)):** \[ D = b^2 - 4ac = 4^2 - 4 \times 3 \times (-32) = 16 + 384 = 400 \] **Find the Square Root of the Discriminant:** \[ \sqrt{D} = \sqrt{400} = 20 \] **Compute the Roots:** \[ y = \frac{-4 \pm 20}{2 \times 3} = \frac{-4 \pm 20}{6} \] This gives us two roots: 1. **First Root (\( y_1 \)):** \[ y_1 = \frac{-4 + 20}{6} = \frac{16}{6} = \frac{8}{3} \] 2. **Second Root (\( y_2 \)):** \[ y_2 = \frac{-4 - 20}{6} = \frac{-24}{6} = -4 \] So, the roots are \( y = -4 \) and \( y = \frac{8}{3} \). --- ### 2. Determine the Intervals Defined by the Roots The roots divide the real number line into three intervals: 1. \( y < -4 \) 2. \( -4 < y < \frac{8}{3} \) 3. \( y > \frac{8}{3} \) --- ### 3. Test Each Interval to See Where the Inequality Holds To determine where \( 3y^{2} + 4y - 32 > 0 \), we'll test a value from each interval. 1. **Interval \( y < -4 \):** - **Test Point:** \( y = -5 \) - **Compute:** \( 3(-5)^2 + 4(-5) - 32 = 3(25) - 20 - 32 = 75 - 20 - 32 = 23 \) - **Result:** \( 23 > 0 \) (Inequality holds) 2. **Interval \( -4 < y < \frac{8}{3} \):** - **Test Point:** \( y = 0 \) - **Compute:** \( 3(0)^2 + 4(0) - 32 = 0 + 0 - 32 = -32 \) - **Result:** \( -32 < 0 \) (Inequality does not hold) 3. **Interval \( y > \frac{8}{3} \):** - **Test Point:** \( y = 3 \) - **Compute:** \( 3(3)^2 + 4(3) - 32 = 3(9) + 12 - 32 = 27 + 12 - 32 = 7 \) - **Result:** \( 7 > 0 \) (Inequality holds) --- ### **Conclusion** The inequality \( 3y^{2} + 4y - 32 > 0 \) holds true in the intervals where \( y < -4 \) or \( y > \frac{8}{3} \). **Solution Set:** \[ y \in (-\infty, -4) \cup \left( \frac{8}{3}, \infty \right) \] --- **Graphical Representation:** A parabola representing \( 3y^{2} + 4y - 32 \) opens upwards (since the coefficient of \( y^2 \) is positive). It crosses the y-axis at the roots \( y = -4 \) and \( y = \frac{8}{3} \). The regions where the parabola is above the x-axis correspond to the solution of the inequality.