0.4 Solving quadratic inequalities

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Introduction

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

$\begin{array}{c}\hfill a{x}^{2}+bx+c>0\\ \hfill a{x}^{2}+bx+c\ge 0\\ \hfill a{x}^{2}+bx+c<0\\ \hfill a{x}^{2}+bx+c\le 0\end{array}$

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.

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

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

4. 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 ${x}^{2}-5x+6\ge 0$ .

1. The factors of ${x}^{2}-5x+6$ are $\left(x-3\right)\left(x-2\right)$ .

2. $\begin{array}{ccc}\hfill {x}^{2}-5x+6& \ge & 0\hfill \\ \hfill \left(x-3\right)\left(x-2\right)& \ge & 0\hfill \end{array}$
3. We need to figure out which values of $x$ satisfy the inequality. From the answers we have five regions to consider.

4. 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 $f\left(2,5\right)=-2,5$ - 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 $x\ge 3$ .

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

Solve the quadratic inequality $-{x}^{2}-3x+5>0$ .

1. 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

$\begin{array}{ccc}\hfill -{x}^{2}-3x+5& =& 0\hfill \\ \hfill {x}^{2}+3x-5& =& 0\hfill \\ \hfill \therefore x& =& \frac{-3±\sqrt{{\left(3\right)}^{2}-4\left(1\right)\left(-5\right)}}{2\left(1\right)}\hfill \\ & =& \frac{-3±\sqrt{29}}{2}\hfill \\ \hfill {x}_{1}& =& \frac{-3-\sqrt{29}}{2}\hfill \\ \hfill {x}_{2}& =& \frac{-3+\sqrt{29}}{2}\hfill \end{array}$
2. We need to figure out which values of $x$ satisfy the inequality. From the answers we have five regions to consider.

3. 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}

4. $-{x}^{2}-3x+5>0$ for ${x}_{1}

When working with an inequality where the variable is in the denominator, a different approach is needed.

Solve $\frac{2}{x+3}\le \frac{1}{x-3}$

1. $\frac{2}{x+3}-\frac{1}{x-3}\le 0$
2. $\begin{array}{c}\hfill \frac{2\left(x-3\right)-\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\le 0\\ \hfill \frac{x-9}{\left(x+3\right)\left(x-3\right)}\le 0\end{array}$
3. We see that the expression is negative for $x<-3$ or $3 .

4. $x<-3\phantom{\rule{1.em}{0ex}}or\phantom{\rule{1.em}{0ex}}3

End of chapter exercises

Solve the following inequalities and show your answer on a number line.

1. Solve: ${x}^{2}-x<12$ .
2. Solve: $3{x}^{2}>-x+4$
3. Solve: ${y}^{2}<-y-2$
4. Solve: $-{t}^{2}+2t>-3$
5. Solve: ${s}^{2}-4s>-6$
6. Solve: $0\ge 7{x}^{2}-x+8$
7. Solve: $0\ge -4{x}^{2}-x$
8. Solve: $0\ge 6{x}^{2}$
9. Solve: $2{x}^{2}+x+6\le 0$
10. Solve for $x$ if: $\frac{x}{x-3}<2$ and $x\ne 3$ .
11. Solve for $x$ if: $\frac{4}{x-3}\le 1$ .
12. Solve for $x$ if: $\frac{4}{{\left(x-3\right)}^{2}}<1$ .
13. Solve for $x$ : $\frac{2x-2}{x-3}>3$
14. Solve for $x$ : $\frac{-3}{\left(x-3\right)\left(x+1\right)}<0$
15. Solve: ${\left(2x-3\right)}^{2}<4$
16. Solve: $2x\le \frac{15-x}{x}$
17. Solve for $x$ :     $\frac{{x}^{2}+3}{3x-2}\le 0$
18. Solve: $x-2\ge \frac{3}{x}$
19. Solve for $x$ : $\frac{{x}^{2}+3x-4}{5+{x}^{4}}\le 0$
20. Determine all real solutions: $\frac{x-2}{3-x}\ge 1$

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