Basic functions: Polynomial functions
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.
\(x<-3 \lor x > 1\)
We have the inequality \[x^2-3 > -2 x\] but first we solve the following equation: \(x^2-3 = -2 x \), that is \( x^2+2 x-3 =0\). We do this via the quadratic formula: \[\begin{aligned} x&=\frac{-2\pm \sqrt{(-2)^2-4 \cdot 1 \cdot -3}}{2}\\ \\ &=\frac{-2\pm \sqrt{16}}{2}\\ \\ &=\frac{-2\pm 4}{2}\end{aligned}\] So \[x=-3\quad \text{or}\quad x=1\] Now we explore where the inequality is true.
First we take a value \(x<-3\), say \(x=-5\). The value of the left-hand side of the inequality is then \[(-5)^2-3=22\] The value of the right-hand side is \[-2 \cdot -5=10\] So we have found for \(x<-3\) that \(x^2-3 > -2 x\).
Next we choose a value \(-3<x<1\), say \(x=-2\). The value of the left-hand side of the inequality is then \[(-2)^2-3=1\] The value of the right-handside is \[-2\cdot -2=4\] So we have found for \(-3<x<1\) that \(x^2-3 < -2 x\).
Finally we choose a value \(x>1\), say \(x=2\). The value of the left-hand side of the inequality is then \[(2)^2-3=1\] The value of the right-hand side is \[-2 \cdot 2=-4\] So we have found for \(x>1\) that \(x^2-3>-2 x\).
So we can conclude that \[x^2-3 > -2 x\] when \(x<-3\) or \(x>1\).