Solution - Solving quadratic inequalities using the quadratic formula

The coefficients of our inequality, $$3x^2+5x-4\le 0$$, are:

$$a$$ = 3

$$b$$ = 5

$$c$$ = -4

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

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

$$x=\left(-5\pm sqrt\left(5^2-4*3*-4\right)\right)/\left(2*3\right)$$

$$x=\left(-5\pm sqrt\left(25-4*3*-4\right)\right)/\left(2*3\right)$$

$$x=\left(-5\pm sqrt\left(25-12*-4\right)\right)/\left(2*3\right)$$

$$x=\left(-5\pm sqrt\left(25--48\right)\right)/\left(2*3\right)$$

$$x=\left(-5\pm sqrt\left(25+48\right)\right)/\left(2*3\right)$$

$$x=\left(-5\pm sqrt\left(73\right)\right)/\left(2*3\right)$$

$$x=\left(-5\pm sqrt\left(73\right)\right)/\left(6\right)$$

to get the result:

$$x=\left(-5\pm sqrt\left(73\right)\right)/6$$

Simplify $$73$$ by finding its prime factors:

The prime factorization of $$73$$ is $$73$$

$$x=\left(-5\pm sqrt\left(73\right)\right)/6$$

The ± means two roots are possible.

Separate the equations:
$$x_1=\left(-5+sqrt\left(73\right)\right)/6$$ and $$x_2=\left(-5-sqrt\left(73\right)\right)/6$$

$$x_1=\left(-5+sqrt\left(73\right)\right)/6$$

$$x_1=\left(-5+sqrt\left(73\right)\right)/6$$

$$x_1=\left(-5+8.544\right)/6$$

$$x_1=\left(-5+8.544\right)/6$$

$$x_1=\left(3.544\right)/6$$

$$x_1=\frac{3.544}{6}$$

$$x_1=0.591$$

$$x_2=\left(-5-sqrt\left(73\right)\right)/6$$

$$x_2=\left(-5-8.544\right)/6$$

$$x_2=\left(-5-8.544\right)/6$$

$$x_2=\left(-13.544\right)/6$$

$$x_2=-\frac{13.544}{6}$$

$$x_2=-2.257$$

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: -2.257, 0.591.

Since the $$a$$ coefficient is positive ($$a$$=3), 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 $$3x^2+5x-4\le 0$$ has a $$\le$$ inequality sign, we look for the parabola intervals that are below the x-axis.

Solution: $$$$

Interval notation: $$$$