# How do you solve the inequality 2x^2+9x+3<=0?

Dec 22, 2016

$x = \frac{- 9 \pm \sqrt{57}}{4}$

$x \approx - 0.362541$
$x \approx - 4.1374586$

#### Explanation:

graph{2x^2+9x+3 [-10, 10, -5, 5]}

I personally like to visualize the inequality, and think "where is the graph of $2 {x}^{2} + 9 x + 3$ less than or equal to zero". So we can set the left side equal to 0, and the solution will be the closed interval between these values.

$2 {x}^{2} + 9 x + 3 \le 0$
$2 {x}^{2} + 9 x + 3 = 0$

$x = \frac{- 9 \pm \sqrt{{9}^{2} - 4 \left(2\right) \left(3\right)}}{2 \left(2\right)}$

$x = \frac{- 9 \pm \sqrt{57}}{4}$

$x \approx - 0.362541$
$x \approx - 4.1374586$

Dec 22, 2016

$x \in \left(\frac{1}{4} \left(- \sqrt{57} - 9\right) , \frac{1}{4} \left(\sqrt{57} - 9\right)\right)$
$= \left(- 4.137 , - 0.3625\right)$, nearly. The segment of the x-axis in the shaded region of the graph illustrates the solution

#### Explanation:

Reorganizing,

${\left(x + \frac{9}{4}\right)}^{2} + \frac{3}{2} - \frac{81}{16} < 0$. So,

$| x + \frac{9}{4} | < \frac{\sqrt{57}}{4}$. And so,

$\frac{1}{4} \left(- \sqrt{57} - 9\right) < x < \frac{1}{4} \left(\sqrt{57} - 9\right)$

graph{2x^2+9x+3 < 0 [-10, 10, -5, 5]}

Dec 22, 2016

interval [-0.414, -0.36]

#### Explanation:

First solve this quadratic equation and find its 2 real roots by using the improved quadratic formula (Socratic Search);
$f \left(x\right) = 2 {x}^{2} + 9 x + 3 \le 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 81 - 24 = 57$ --> $d = \pm \sqrt{57} = \pm 7.55$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{9}{4} \pm \frac{7.55}{4}$
$x 1 = - \frac{16.55}{4} = - 4.14$
$x 2 = - \frac{1.45}{4} = - 0.36$.
Since a = 2 > 0, the parabola graph opens upward. Between the 2 real roots, f(x) < 0, as one part of the graph stays below the x-axis.
There for f(x) < 0 between the 2 real roots.
Answer by closed interval: [- 4.14, 0.36]
Both end points (-0.36 and -4.14) are included in the solution set.
Graph on the Number Line:

-------------------- -4.14 ============= -0.36 ---- 0 ----------------------