Solving quadratic 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.
\(x<-1 \lor x > 5\)
We have the inequality \[x^2-5 > 4 x\] but first we solve the following equation: \(x^2-5 = 4 x \), that is \( x^2-4 x-5 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{4\pm \sqrt{(4)^2-4 \cdot 1 \cdot -5}}{2}\\ \\ &=\frac{4\pm \sqrt{36}}{2}\\ \\ &=\frac{4\pm 6}{2}\end{aligned}\] So \[x=-1\quad \text{or}\quad x=5\] Now we explore where the inequality is true.
First we take a value \(x<-1\), say \(x=-3\). The value of the left-hand side of the inequality is then \[(-3)^2-5=4\] The value of the right-hand side is \[4 \cdot -3=-12\] So we have found for \(x<-1\) that \(x^2-5 > 4 x\).
Next we choose a value \(-1<x<5\), say \(x=0\). The value of the left-hand side of the inequality is then \[(0)^2-5=-5\] The value of the right-handside is \[4\cdot 0=0\] So we have found for \(-1<x<5\) that \(x^2-5 < 4 x\).
Finally we choose a value \(x>5\), say \(x=6\). The value of the left-hand side of the inequality is then \[(6)^2-5=31\] The value of the right-hand side is \[4 \cdot 6=24\] So we have found for \(x>5\) that \(x^2-5>4 x\).
So we can conclude that \[x^2-5 > 4 x\] when \(x<-1\) or \(x>5\).