# Quadratic Inequalities - Examples

**Solved Example 1:** Solve the inequality .

**Solution:** The expression \(2{x^2} + 5x + 2\) can be factorized \(\left( {2x + 1} \right)\left( {x + 2} \right)\), so that its zeroes are \(x = - \frac{1}{2}\) and \(x = - 2\). Thus, \(2{x^2} + 5x + 2 > 0\) when

\[x < - 2\,\,\,{\rm{or}}\,\,\,x > - \frac{1}{2}\]

**Solved Example 2:** Solve the inequality \(6{x^2} - 5x + 1 < 0\).

**Solution:** We have:

\[\begin{array}{l}\left( {3x - 1} \right)\left( {2x - 1} \right) < 0\\ \Rightarrow \,\,\,\frac{1}{3} < x < \frac{1}{2}\end{array}\]

**Solved Example 3:** Solve the inequality \( - {x^2} - 4x - 3 > 0\).

**Solution:** We have:

\[\begin{array}{l} - \left( {{x^2} + 4x + 3} \right) > 0\\ \Rightarrow \,\,\,{x^2} + 4x + 3 < 0\\ \Rightarrow \,\,\,\left( {x + 1} \right)\left( {x + 3} \right) < 0\\ \Rightarrow \,\,\, - 3 < x < - 1\end{array}\]

What if the quadratic expression in an inequality has no real zeroes? Suppose that you have to find the values of *x* which satisfy \({x^2} + x + 1 > 0\). The expression \({x^2} + x + 1\) has no real zeroes. If you plot its graph, you will obtain a parabola which lies entirely above the horizontal axis:

It is easy to see that \({x^2} + x + 1 > 0\) for *all values of* *x*. Thus, the solution set of \({x^2} + x + 1 > 0\) is the set of all real numbers.

On the other hand, the solution set of \({x^2} + x + 1 < 0\) will be empty, because the value of \({x^2} + x + 1\) is *never* negative.

**Solved Example 4:** Solve the inequality \({x^2} + 2x + 3 < 0\).

**Solution:** We note that the discriminant for the quadratic is \(D = {2^2} - 4 \times 3 = - 8 < 0\), so the graph of the quadratic expression \({x^2} + 2x + 3\) never intersects the horizontal axis – in fact, it always lies above the *x*-axis. Thus, \({x^2} + 2x + 3 < 0\) is satisfied for no real value of *x*.

**Solved Example 5:** Solve the inequality \( - 1 - 2x - 3{x^2} > 0\).

**Solution:** We can rewrite the given inequality as follows:

\[\begin{array}{l} - \left( {3{x^2} + 2x + 1} \right) > 0\\ \Rightarrow \,\,\,3{x^2} + 2x + 1 < 0\end{array}\]

We note that the discrminant of the quadratic expression \(3{x^2} + 2x + 1\) is \(D = {2^2} - 4 \times 3 = - 8 < 0\), so the corresponding parabola will always lie above the horizontal axis. Thus, \(3{x^2} + 2x + 1 < 0\) is satisfied for no real value of *x*.

**Solved Example 6:** Let \(f\left( x \right)\) be a quadratic expressions. Suppose that the value of this expression is evaluated at three different points: \(x = a,\,\,x = b,\,\,x = c\), where \(a < b < c\). It is found that

\[f\left( a \right) > 0,\,f\left( b \right) < 0,\,\,f\left( c \right) > 0\]

What can you say about the zeroes of \(f\left( x \right)\)?

**Solution:** This problem can be solved without any calculations. Now that since \(f\left( a \right) > 0\) and \(f\left( b \right) < 0\), the parabola of the expression must lie above the *x*-axis for \(x = a\) and below that *x*-axis for \(x = b\). This means the parabola must cross the *x*-axis between \(x = a\) and \(x = b\) (as indicated by the heavy point):

Similarly, \(f\left( b \right) < 0\) and \(f\left( c \right) > 0\), the parabola must cross the axis again between and :

Thus, not only can we say that \(f\left( x \right)\) has real and distinct zeroes, we can also specify the location of these zeroes: one zero is in the interval \(\left( {a,b} \right)\) and the second zero is in the interval \(\left( {b,c} \right)\). This graphical approach to problem solving in quadratics is extremely powerful!