Quadratic Inequalities - Theory

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Suppose that you are asked to find out all the values of x for which \({x^2} - 3x + 2 > 0\). Let us first try an algebraic approach. By factorization, we can write the given inequality as

\[\left( {x - 1} \right)\left( {x - 2} \right) > 0\]

The product of two terms will be positive if both of them are positive, or both are negative. The two brackets on the left side above are both positive if \(x > 2\), and are both negative if \(x < 1\)\( - {x^2} + 3x - 2 < 0\). Thus, we can say that values of x which satisfy this inequality are \(x < 1\) or \(x > 2\).

A graphical approach is also possible. The graph of \(Q\left( x \right)\,\,:\,\,{x^2} - 3x + 2\) is plotted below:

Note the heavy parts of the curve. These lie above the horizontal axis, that is, they correspond to positive values of the expression. The corresponding x-values are \(x < 1\) or \(x > 2\).

Now, consider the quadratic inequality \({x^2} - 3x + 2 < 0\). This can be written as

\[\left( {x - 1} \right)\left( {x - 2} \right) < 0\]

The product of two terms is negative if the two terms are of opposite signs. The two brackets on the left side above are of opposite signs if x is less than 2 but greater than 1. Thus, this inequality is satisfied if \(1 < x < 2\). Graphically, this corresponds to the part of the parabola below the -axis:

Next, consider the inequality . By multiplying with \( - 1\) across both sides of the inequality, it changes to:

\[{x^2} - 3x + 2 > 0\]

Note that the sign of the inequality has reversed, because we multiplied with a negative quantity on both sides. Now, it can be solved as done earlier to obtain the solution as \(x < 1\) or \(x > 2\).

Alternatively, if we plot the graph of \(Q\left( x \right)\,\,:\,\, - {x^2} + 3x - 2\), we will obtain an inverted parabola, as shown below. We note that \(Q\left( x \right) > 0\) when \(x < 1\) or \(x > 2\). This is the same answer we obtained algebraically.

To summarize, suppose that we have a quadratic expression \(Q\left( x \right)\,\,:\,\,a{x^2} + bx + c\) whose zeroes are \(\alpha ,\,\,\beta \).

If \(a > 0\), then

\[\begin{array}{l} Q\left( x \right) > 0\,\,{\rm{when}}\,\,x < \alpha \,\,{\rm{or}}\,\,x > \beta \\Q\left( x \right) < 0\,\,{\rm{when}}\,\,\alpha < x < \beta \end{array}\]

If \(a < 0\), then

\[\begin{array}{l}Q\left( x \right) > 0\,\,{\rm{when}}\,\,\alpha  < x < \beta \\Q\left( x \right) < 0\,\,{\rm{when}}\,\,x < \alpha \,\,{\rm{or}}\,\,x > \beta \end{array}\]

Remembering the following diagrams will help: