# Inequalities – Finding when a line in below a parabola – Exercise 5719

Exercise

Find the section in x-aixs where the line:

$$y=x+1$$

is below the parabla:

$$y=x^2-2x-9$$

$$x<-2 \text{ or } x>5$$

Solution

We want to find for which points in x-aixs the following inequality is true:

$$x^2-2x-9>x+1$$

Move everything to one side:

$$x^2-2x-9-x-1>0$$

$$x^2-3x-10>0$$

It is a square inequality. Its coefficients are

$$a=1, b=-3, c=-10$$

The coefficient of the square expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is above the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{3\pm \sqrt{{(-3)}^2-4\cdot 1\cdot (-10)}}{2\cdot 1}=$$

$$=\frac{3\pm \sqrt{49}}{2}=$$

$$=\frac{3\pm 7}{2}$$

Hence, we get the solutions:

$$x_1=\frac{3+ 7}{2}=5$$

$$x_2=\frac{3- 7}{2}=-2$$

Because the parabola we got “smiles” and we look for the section where it is above the x-axis, we get that the requested section is

$$x<-2 \text{ or } x>5$$

Here are the line and the parabola:

You can see that the line is indeed below the parabola in the section we found.

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