Questions: Solve the given polynomial inequality. Report the solution in both interval notation and graphical form. -x^2+x+72<0

Solution Steps

The given inequality is: \[ -x^2 + x + 72 < 0 \]

To find the roots, we solve the equation: \[ -x^2 + x + 72 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 1 \), and \( c = 72 \): \[ x = \frac{-1 \pm \sqrt{1^2 - 4(-1)(72)}}{2(-1)} \] \[ x = \frac{-1 \pm \sqrt{1 + 288}}{-2} \] \[ x = \frac{-1 \pm \sqrt{289}}{-2} \] \[ x = \frac{-1 \pm 17}{-2} \]

So, the roots are: \[ x = \frac{-1 + 17}{-2} = -8 \] \[ x = \frac{-1 - 17}{-2} = 9 \]

The roots divide the number line into three intervals: \( (-\infty, -8) \), \( (-8, 9) \), and \( (9, \infty) \).

We test a point in each interval to determine where the inequality holds.

1. For \( x = -9 \) in \( (-\infty, -8) \): \[ -(-9)^2 + (-9) + 72 = -81 - 9 + 72 = -18 \] \[ -18 < 0 \] (True)

2. For \( x = 0 \) in \( (-8, 9) \): \[ -(0)^2 + 0 + 72 = 72 \] \[ 72 < 0 \] (False)

3. For \( x = 10 \) in \( (9, \infty) \): \[ -(10)^2 + 10 + 72 = -100 + 10 + 72 = -18 \] \[ -18 < 0 \] (True)

Final Answer