x^2+4x+13< 0
Gotta solve the inequation.
Hello,
rearrange:
$\displaystyle x^2+4x+13<0~\implies~x^2+4x+4<-13+4~\implies~(x+2)^2<-9$
A square can't never be negative, thus there doesn't exist a real solution.
If there is a typo and the original inequality reads:
$\displaystyle x^2+4x+3<0~\implies~x^2+4x+4<-3+4~\implies~(x+2)^2<1$ then you get:
$\displaystyle -1 < x+2 < 1~\implies~-3 < x < -1$
Hello, nerdzor!
Another approach . . .
Solve: .$\displaystyle x^2 + 4x + 13 \:< \: 0$
We have the function: .$\displaystyle y \:=\:x^2 + 4x + 13$ . . . a parabola
. . and they want to know when the graph is below the x-axis.
Since this is an up-opening parabola, the curve is below the x-axis
. . in the interval between the x-intercepts.
The x-intercepts occur when $\displaystyle y = 0\!:\;\;x^2 + 4x + 13 \:=\:0$
. . But this equation has no real roots.
Hence, there are no x-intercepts. .The graph never goes below the x-axis.
The inequality is never true.