Solution - Solving quadratic inequalities using the quadratic formula

To find the roots of a quadratic equation, plug its coefficients ($$a$$, $$b$$ and $$c$$ ) into the quadratic formula:

$$x=\left(-b\pm sqrt\left(b^2-4ac\right)\right)/\left(2a\right)$$

$$x=\left(-1*-13\pm sqrt\left(-{13}^2-4*1*-48\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-13\pm sqrt\left(169-4*1*-48\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-13\pm sqrt\left(169-4*-48\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-13\pm sqrt\left(169--192\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-13\pm sqrt\left(169+192\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-13\pm sqrt\left(361\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-13\pm sqrt\left(361\right)\right)/\left(2\right)$$

$$x=\left(13\pm sqrt\left(361\right)\right)/2$$

to get the result:

$$x=\left(13\pm sqrt\left(361\right)\right)/2$$

Simplify $$361$$ by finding its prime factors:

The prime factorization of $$361$$ is $${19}^2$$

$$x=\left(13\pm 19\right)/2$$

The ± means two roots are possible.

Separate the equations:
$$x_1=\left(13+19\right)/2$$ and $$x_2=\left(13-19\right)/2$$

$$x_1=\left(13+19\right)/2$$

$$x_1=\left(13+19\right)/2$$

$$x_1=\left(32\right)/2$$

$$x_1=\frac{32}{2}$$

$$x_1=16$$

$$x_2=\left(13-19\right)/2$$

$$x_2=\left(13-19\right)/2$$

$$x_2=\left(-6\right)/2$$

$$x_2=-\frac{6}{2}$$

$$x_2=-3$$

To find the intervals of a quadratic inequality, we start by finding its parabola.

The roots of the parabola (where it meets the x-axis) are: -3, 16.

Since the $$a$$ coefficient is positive ($$a$$=1), this is a "positive" quadratic inequality and the parabola points upward, like a smile!

If the inequality sign is ≤ or ≥ , then the intervals include the roots and we use a solid line. If the inequality sign is < or > the intervals do not include the roots and we use a dotted line.

Since $$x^2-13x-48<0$$ has a $$<$$ inequality sign, we look for the parabola intervals that are below the x-axis.

Solution: $$$$

Interval notation: $$$$