In order to solve this inequality, we have to use the definition of the absolute value function, that is:
|f(x)| > a ⇒ f(x) > a , f(x) < -a
Applying this to our expression, we get:
[tex]\begin{gathered} \lvert x^2+3x-29\rvert>25 \\ x^2+3x-29>25,x^2+3x-29<-25 \end{gathered}[/tex]
We got two inequalities:
1) x^2 + 3x - 29 > 25
2) x^2 + 3x - 29 < -25
Let's solve the first one, first, we have to rewrite the equation so there is a 0 on the right side:
x^2 + 3x - 29 > 25
x^2 + 3x - 29 - 25 > 25 - 25
x^2 + 3x - 54 > 0
In order to solve this, we can apply the quadratic equation, like this:
[tex]\begin{gathered} \frac{-b\pm\sqrt[]{b^2-4ac}}{2c} \\ \frac{-3\pm\sqrt[]{(3)^2-4\times1\times(-54)}}{2\times1}=\frac{-3\pm15}{2} \\ x1=\frac{-3+15}{2}=\frac{12}{2}=6 \\ x2=\frac{-3-15}{2}=\frac{-18}{2}=-9 \end{gathered}[/tex]
Since a>0 (1, the number that multiplies x^2) then the parabola opens upwards, then the above inequality is true if x<-9 or x>6.
Now, let's solve the second inequality, similarly as for the previous one, we get:
x^2 + 3x - 29 < -25
x^2 + 3x - 4 < 0
Applying the quadratic formula, we get the solutions of this inequality, x1 = 1 and x2= -4.
Then, the inequality is true if x<1 and x>-4 (-4
Mixing the two domains:
-46)
x<-9 or -46
As you can see, the domain has 3 different areas, from -∞ to -9 (-∞, -9), from -4 to 1 (-4, 1) and from 6 to ∞ (6, ∞), then the solution set in interval notation is:
(-∞, -9), (-4, 1), (6, ∞).
We can graph this solution set on a real number line, like this:
