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Now that you know how to solve quadratic equations, you are ready to learn how to solve quadratic inequalities.
A quadratic inequality is an inequality of the form
Solving a quadratic inequality corresponds to working out in what region the graph of a quadratic function lies above or below the $x$ -axis.
Solve the inequality $4{x}^{2}-4x+1\le 0$ and interpret the solution graphically.
Let $f\left(x\right)=4{x}^{2}-4x+1$ . 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 $f\left(x\right)=4{x}^{2}-4x+1$ touches the $x$ -axis at $x=\frac{1}{2}$ , but there are no regions where the graph is below the $x$ -axis.
Find all the solutions to the inequality ${x}^{2}-5x+6\ge 0$ .
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.
$f\left(x\right)$ | sign of $f\left(x\right)$ | ||
Region A | $x<2$ | $f\left(1\right)=2$ | + |
Region B | $x=2$ | $f\left(2\right)=0$ | + |
Region C | $2<x<3$ | $f(2,5)=-2,5$ | - |
Region D | $x=3$ | $f\left(3\right)=0$ | + |
Region E | $x>3$ | $f\left(4\right)=2$ | + |
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 $x\ge 3$ .
Solve the quadratic inequality $-{x}^{2}-3x+5>0$ .
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 ${x}_{1}<x<{x}_{2}$
When working with an inequality where the variable is in the denominator, a different approach is needed.
Solve $\frac{2}{x+3}}\le {\displaystyle \frac{1}{x-3}$
We see that the expression is negative for $x<-3$ or $3<x\le 9$ .
Solve the following inequalities and show your answer on a number line.
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