# X^2+8x+1<0 Inequality questions What will be the answer ? Plz tell

Jun 14, 2018

$- 4 - \sqrt{15} < x < - 4 + \sqrt{15}$

#### Explanation:

Complete the square:
${x}^{2} + 8 x + 1 < 0$
${\left(x + 4\right)}^{2} - 15 < 0$
${\left(x + 4\right)}^{2} < 15$
$| x + 4 | < \sqrt{15}$

If $x + 4 \ge 0$, then $x < - 4 + \sqrt{15}$.
If $x + 4 < 0$, then $- x - 4 < \sqrt{15} \Rightarrow x > - 4 - \sqrt{15}$

So we have two ranges for $x$:
$- 4 \le x < - 4 + \sqrt{15}$ and $- 4 - \sqrt{15} < x < - 4$.
We can combine these to make one range:
$- 4 - \sqrt{15} < x < - 4 + \sqrt{15}$

Numerically, to three significant figures:
$- 7.87 < x < - 0.127$

Jun 14, 2018

$\left(- 4 - \sqrt{15} , - 4 + \sqrt{15}\right)$

#### Explanation:

$f \left(x\right) = {x}^{2} + 8 x + 1 < 0$
First, solve the quadratic equation f(x) = 0, to find the 2 end-points (critical points).
$D = {d}^{2} = {b}^{2} - 4 a c = 64 - 4 = 60$ --> $d = \pm 2 \sqrt{15}$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{8}{2} \pm 2 \frac{\sqrt{15}}{2} = - 4 \pm \sqrt{15}$
$x 1 = - 4 - \sqrt{15}$, and x2 = - 4 + sqrt15).
The graph of f(x) is a upward parabola (a > 0). Between the 2 real roots (x1 , x2), the graph is below the x-axis --> f (x) < 0.
The answer is the open interval:
$\left(- 4 - \sqrt{15} , - 4 + \sqrt{15}\right)$