1. ## Year 10 Quadratic inequalities

x^2+4x+13< 0

Gotta solve the inequation.

2. Originally Posted by nerdzor
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:

$\displaystyle -1 < x+2 < 1~\implies~-3 < x < -1$

3. 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.