Solving equations and inequalities: Quadratic inequalities
Solving quadratic inequalities via the quadratic formula and inspection
- first solving the corresponding quadratic equation;
- then figuring out in which area(s) the inequality is true;
- finally, combining the intermediate results.
\(x<-4 \lor x > 1\)
We have the inequality \[x^2-4 > -3 x\] but first we solve the following equation: \(x^2-4 = -3 x \), that is \( x^2+3 x-4 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{-3\pm \sqrt{(-3)^2-4 \cdot 1 \cdot -4}}{2}\\ \\ &=\frac{-3\pm \sqrt{25}}{2}\\ \\ &=\frac{-3\pm 5}{2}\end{aligned}\] So \[x=-4\quad \text{or}\quad x=1\] Now we explore where the inequality is true.
First we take a value \(x<-4\), say \(x=-6\). The value of the left-hand side of the inequality is then \[(-6)^2-4=32\] The value of the right-hand side is \[-3 \cdot -6=18\] So we have found for \(x<-4\) that \(x^2-4 > -3 x\).
Next we choose a value \(-4<x<1\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-4=5\] The value of the right-handside is \[-3\cdot -3=9\] So we have found for \(-4<x<1\) that \(x^2-4 < -3 x\).
Finally we choose a value \(x>1\), say \(x=2\). The value of the left-hand side of the inequality is then \[(2)^2-4=0\] The value of the right-hand side is \[-3 \cdot 2=-6\] So we have found for \(x>1\) that \(x^2-4>-3 x\).
So we can conclude that \[x^2-4 > -3 x\] when \(x<-4\) or \(x>1\).