Quadratic Inequalities (DP IB Analysis & Approaches (AA): HL): Revision Note

Quadratic inequalities

What affects the inequality sign when rearranging a quadratic inequality?

• The inequality sign is unchanged by

  • Adding or subtracting a term to both sides

    • E.g. $\begin{array}{rcl}x& >& 9\end{array}$ becomes $\begin{array}{rcl}x+3& >& 12\end{array}$

  • Multiplying or dividing both sides by a positive term

    • E.g. $\begin{array}{rcl}x& >& 9\end{array}$ becomes $\begin{array}{rcl}3x& >& 27\end{array}$

• The inequality sign flips (< changes to >) when

  • Multiplying or dividing both sides by a negative term

    • E.g. $\begin{array}{rcl}x& >& 9\end{array}$ becomes $\begin{array}{rcl}-3x& <& -27\end{array}$

How do I solve a quadratic inequality?

• STEP 1

  Rearrange the inequality into quadratic form with a positive squared term and with zero on one side

  • E.g. $a{x}^2+bx+c>0$

• STEP 2

  Find the roots of the quadratic equation

  • Solve $a{x}^2+bx+c=0$ to get $x_1$ and $x_2$ where $x_1<x_2$

• STEP 3
  Sketch a graph of the quadratic and label the roots

  • As the squared term is positive it will be concave up, i.e. U-shaped

• STEP 4
  Identify the region on the graph which satisfies the inequality

  • If you want the graph to be above the $x$-axis then the region will be the two intervals outside of the two roots

    • E.g. for $a{x}^2+bx+c>0$, the solution is $x<x_1$ or $x>x_2$

    • and for $a{x}^2+bx+c\ge 0$, the solution is $x\le x_1$ or $x\ge x_2$

  • If you want the graph to be below the $x$-axis then choose the region to be the interval between the two roots

    • E.g. for $a{x}^2+bx+c<0$, the solution is $x_1<x<x_2$

    • and for $a{x}^2+bx+c\le 0$, the solution is $x_1\le x\le x_2$

How do I solve a quadratic inequality of the form (x-h)²<n or (x-h)²>n?

• The safest way is to expand and rearrange, then follow the steps above

• A common mistake is writing $x-h<\pm \sqrt{n}$ or $x-h>\pm \sqrt{n}$

  • This is NOT correct!

• The correct solution to${(x-h)}^2<n$ is

  • $\left|x-h\right|<\sqrt{n}$ which can be written as $-\sqrt{n}<x-h<\sqrt{n}$

  • The final solution is $h-\sqrt{n}<x<h+\sqrt{n}$

• The correct solution to ${(x-h)}^2>n$ is

  • $\left|x-h\right|>\sqrt{n}$ which can be written as $x-h<-\sqrt{n}$ or $x-h>\sqrt{n}$

  • The final solution is $x<h-\sqrt{n}$ or $x>h+\sqrt{n}$