<p>A quadratic inequality has two critical values. Solve it by sketching the parabola and reading off where the curve lies above or below the \(x\)-axis.</p>
<div class="titled-block"><div class="titled-block-title">Method: Arrange, Factorise, Sketch</div><ol class="num-list"><li data-list-item-id="u01-t2-l1" data-list-item-path="u01-t2-l1"><div class="list-item-paragraph">Arrange all terms on one side so that the other side is zero.</div></li><li data-list-item-id="u01-t2-l2" data-list-item-path="u01-t2-l2"><div class="list-item-paragraph">If the coefficient of \(x^2\) is negative, multiply both sides by \(-1\) and reverse the inequality sign.</div></li><li data-list-item-id="u01-t2-l3" data-list-item-path="u01-t2-l3"><div class="list-item-paragraph">Factorise to find the critical values.</div></li><li data-list-item-id="u01-t2-l4" data-list-item-path="u01-t2-l4"><div class="list-item-paragraph">Sketch the parabola and identify where it is above or below the \(x\)-axis.</div></li><li data-list-item-id="u01-t2-l5" data-list-item-path="u01-t2-l5"><div class="list-item-paragraph">Write the range of \(x\); include the endpoints only for&nbsp;\(\ge\) or \(\le\).</div></li></ol></div>
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<p><strong>Reading the sketch (roots \(p&lt;q\), coefficient of \(x^2\) positive)</strong></p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 540px;"><colgroup><col style="width: 40.7407%;"><col style="width: 59.2593%;"></colgroup><thead><tr><th style="text-align: center; --cell-justify: center"><strong>Inequality</strong></th><th style="text-align: center; --cell-justify: center"><strong>Solution</strong></th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">\(ax^2+bx+c&gt;0\)</td><td style="text-align: center; --cell-justify: center">\(x&lt;p\) or \(x&gt;q\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(ax^2+bx+c\ge 0\)</td><td style="text-align: center; --cell-justify: center">\(x\le p\) or \(x\ge q\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(ax^2+bx+c&lt;0\)</td><td style="text-align: center; --cell-justify: center">\(p&lt;x&lt;q\)</td></tr><tr><td style="text-align: center; --cell-justify: center">\(ax^2+bx+c\le 0\)</td><td style="text-align: center; --cell-justify: center">\(p\le x\le q\)</td></tr></tbody></table></div></div>
<p><strong>Common mistakes with quadratic inequalities</strong></p>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 600px;"><colgroup><col style="width: 50.0000%;"><col style="width: 50.0000%;"></colgroup><thead><tr><th style="text-align: center; --cell-justify: center"><strong>Wrong</strong></th><th style="text-align: center; --cell-justify: center"><strong>Correct</strong></th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">\[\begin{aligned}x^2&&gt;9\\x&&gt;\pm3\end{aligned}\]</td><td style="text-align: center; --cell-justify: center">\[\begin{aligned}x^2&&gt;9\\x^2-9&&gt;0\\(x+3)(x-3)&&gt;0\\x&&lt;-3\quad\text{or}\quad x&gt;3\end{aligned}\]<br><br>\(\textbf{Note:}\) Do not take square roots directly on both sides of an inequality.</td></tr><tr><td style="text-align: center; --cell-justify: center">\[\begin{aligned}x^2&&gt;6x\\\dfrac{x^2}{x}&&gt;\dfrac{6x}{x}\\x&&gt;6\end{aligned}\]</td><td style="text-align: center; --cell-justify: center">\[\begin{aligned}x^2&&gt;6x\\x^2-6x&&gt;0\\x(x-6)&&gt;0\\x&&lt;0\quad\text{or}\quad x&gt;6\end{aligned}\]<br><br>\(\textbf{Note:}\) Do not divide both sides by \(x\) because its sign is unknown.</td></tr><tr><td style="text-align: center; --cell-justify: center">\[\begin{aligned}(x-2)(x-4)&&lt;0\\x-2&&lt;0\quad\text{or}\quad x-4&gt;0\\x&&lt;2\quad\text{or}\quad x&gt;4\end{aligned}\]</td><td style="text-align: center; --cell-justify: center">\[\begin{aligned}(x-2)(x-4)&&lt;0\\x-2&lt;0,\ x-4&&gt;0\quad\text{or}\quad x-2&gt;0,\ x-4&lt;0\\x&lt;2,\ x&&gt;4\quad\text{or}\quad x&gt;2,\ x&lt;4\\&\text{no solution}\quad\text{or}\quad 2&lt;x&lt;4\\\therefore\quad &2&lt;x&lt;4\end{aligned}\]<br><br>\(\textbf{Note:}\) Do not split a product inequality into separate linear inequalities.</td></tr></tbody></table></div></div>
<div class="content-box-placement content-box-placement--full content-box-placement--align-left" style="display:flex;justify-content:stretch;max-width:100%"><div class="content-box content-box--emphasis" data-content-box-preset="emphasis" style="box-sizing:border-box;border-color:var(--content-box-border, #2563eb);border-style:solid;border-width:2px;border-radius:6px;background-color:var(--content-box-background, #ffffff);color:var(--content-box-foreground, #111827);padding:14px 18px;font-family:'EB Garamond', serif;font-size:16px;line-height:1.35;width:100%;max-width:100%"><div class="content-box-title">Exam check</div><p>Be careful when the coefficient of \(x^2\) is negative: multiplying or dividing an inequality by a negative number reverses the inequality sign.</p></div></div>

A quadratic inequality has two critical values. Solve it by sketching the parabola and reading off where the curve lies above or below the \(x\)-axis.

Arrange all terms on one side so that the other side is zero.

If the coefficient of \(x^2\) is negative, multiply both sides by \(-1\) and reverse the inequality sign.

Sketch the parabola and identify where it is above or below the \(x\)-axis.

Write the range of \(x\); include the endpoints only for \(\ge\) or \(\le\).

Method: Arrange, Factorise, Sketch

**Reading the sketch (roots \(p<q\), coefficient of \(x^2\) positive)**

**Common mistakes with quadratic inequalities**

\[\begin{aligned}x^2&>9\\x&>\pm3\end{aligned}\]

\[\begin{aligned}x^2&>9\\x^2-9&>0\\(x+3)(x-3)&>0\\x&<-3\quad\text{or}\quad x>3\end{aligned}\]

\(\textbf{Note:}\) Do not take square roots directly on both sides of an inequality.

\[\begin{aligned}x^2&>6x\\\dfrac{x^2}{x}&>\dfrac{6x}{x}\\x&>6\end{aligned}\]

\[\begin{aligned}x^2&>6x\\x^2-6x&>0\\x(x-6)&>0\\x&<0\quad\text{or}\quad x>6\end{aligned}\]

\(\textbf{Note:}\) Do not divide both sides by \(x\) because its sign is unknown.

\[\begin{aligned}(x-2)(x-4)&<0\\x-2&<0\quad\text{or}\quad x-4>0\\x&<2\quad\text{or}\quad x>4\end{aligned}\]

\[\begin{aligned}(x-2)(x-4)&<0\\x-2<0,\ x-4&>0\quad\text{or}\quad x-2>0,\ x-4<0\\x<2,\ x&>4\quad\text{or}\quad x>2,\ x<4\\&\text{no solution}\quad\text{or}\quad 2<x<4\\\therefore\quad &2<x<4\end{aligned}\]

\(\textbf{Note:}\) Do not split a product inequality into separate linear inequalities.

Be careful when the coefficient of \(x^2\) is negative: multiplying or dividing an inequality by a negative number reverses the inequality sign.

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