First ignore the inequality sign and solve as a quadratic equation to find the roots. We can factorise harder (non-monic) quadratics like $2 x^{2} + 3 x - 2$ by "grouping" (or by "using a grid" method)

$\begin{array}{rcl} 2 x^{2} + 3 x - 2 & = & 0 \end{array}$

$\begin{array}{rcl} a c & = & 2 \times (- 2) = - 4 \end{array}$ , $\begin{array}{rcl} b & = & 3 \end{array}$ ; two numbers that times to −4 and add to 3 are 4 and −1

$\begin{array}{rcl} 2 x^{2} + 4 x - x - 2 & = & 0 \end{array}$

Group and factorise the first two terms and the last two terms separately

$\begin{array}{rcl} 2 x (x + 2) - (x + 2) & = & 0 \end{array}$

Factorise into final form using the common factor $(x + 2)$

$\begin{array}{rcl} (2 x - 1) (x + 2) & = & 0 \end{array}$

[1]

Solve

$\begin{array}{rcl} 2 x - 1 & = & 0, x + 2 = 0 \\ x & = & \frac{1}{2}, x = - 2 \end{array}$

[1]

Now that we have the roots, we need to consider the original inequality $2 x^{2} + 3 x - 2 > 0$
Sketch a graph of the quadratic. The squared term is positive so we sketch a U-shaped parabola...

edexcel-tq-mis-2-18-h-1-1

...and $" > 0 "$ tells us to look for values above the $x$ -axis. It helps to shade the region we don't want- below the $x$ -axis

edexcel-tq-mis-2-18-h-1-2

$x < - 2, x > \frac{1}{2}$ [1]
Always sketch the graph to decide which way to put the inequality signs- they must be correct to achieve the final mark

$x < - 2.5 or x > 1$	$x > - 2.5 or x > 1$
$- 2.5 < x < 1$	$x > - 2.5 or x < 1$

Solving Inequalities (Edexcel IGCSE Maths)

Exam Questions