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:

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

$$x=\left(-0\pm sqrt\left(0^2-4*1*-16\right)\right)/\left(2*1\right)$$

$$x=\left(-0\pm sqrt\left(0-4*1*-16\right)\right)/\left(2*1\right)$$

$$x=\left(-0\pm sqrt\left(0-4*-16\right)\right)/\left(2*1\right)$$

$$x=\left(-0\pm sqrt\left(0--64\right)\right)/\left(2*1\right)$$

$$x=\left(-0\pm sqrt\left(0+64\right)\right)/\left(2*1\right)$$

$$x=\left(-0\pm sqrt\left(64\right)\right)/\left(2*1\right)$$

$$x=\left(-0\pm sqrt\left(64\right)\right)/\left(2\right)$$

to get the result:

$$x=\left(-0\pm sqrt\left(64\right)\right)/2$$

Simplify $$64$$ by finding its prime factors:

The prime factorization of $$64$$ is $$2^6$$

$$x=\left(-0\pm 8\right)/2$$

The ± means two roots are possible.

Separate the equations:
$$x_1=\left(-0+8\right)/2$$ and $$x_2=\left(-0-8\right)/2$$

$$x_1=\left(-0+8\right)/2$$

$$x_1=\left(-0+8\right)/2$$

$$x_1=\left(8\right)/2$$

$$x_1=\frac{8}{2}$$

$$x_1=4$$

$$x_2=\left(-0-8\right)/2$$

$$x_2=\left(-0-8\right)/2$$

$$x_2=\left(-8\right)/2$$

$$x_2=-\frac{8}{2}$$

$$x_2=-4$$

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: -4, 4.

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 $$x^2+0x-16<0$$ has a $$<$$ inequality sign, we look for the parabola intervals that are below the x-axis.

Solution: $$$$

Interval notation: $$$$