0.4 Solving quadratic inequalities

Let $f\left(x\right)=4x^2-4x+1$ . Factorising this quadratic function gives $f\left(x\right)={(2x-1)}^2$ .

$f\left(x\right)=0$ only when $x=\frac{1}{2}$ .

This means that the graph of $f\left(x\right)=4x^2-4x+1$ touches the $x$ -axis at $x=\frac{1}{2}$ , but there are no regions where the graph is below the $x$ -axis.

The factors of $x^2-5x+6$ are $(x-3)(x-2)$ .

We need to figure out which values of $x$ satisfy the inequality. From the answers we have five regions to consider.

Let $f\left(x\right)=x^2-5x+6$ . For each region, choose any point in the region and evaluate the function.

We see that the function is positive for $x\le 2$ and $x\ge 3$ .

We see that $x^2-5x+6\ge 0$ is true for $x\le 2$ and $x\ge 3$ .

Let $f\left(x\right)=-x^2-3x+5$ . $f\left(x\right)$ cannot be factorised so, use the quadratic formula to determine the roots of $f\left(x\right)$ . The $x$ -intercepts are solutions to the quadratic equation

We need to figure out which values of $x$ satisfy the inequality. From the answers we have five regions to consider.

We can use another method to determine the sign of the function over different regions, by drawing a rough sketch of the graph of the function. We know that the roots of the function correspond to the $x$ -intercepts of the graph. Let $g\left(x\right)=-x^2-3x+5$ . We can see that this is a parabola with a maximum turning point that intersects the $x$ -axis at $x_1$ and $x_2$ .

It is clear that $g\left(x\right)>0$ for $x_1<x<x_2$

$-x^2-3x+5>0$ for $x_1<x<x_2$

We see that the expression is negative for $x<-3$ or $3<x\le 9$ .