Question 1.37: Solve the following inequalities: (a) x²+x-6   (b) x²+8x+1&......

For the product (x — 2)(x + 3) to be positive requires either

(i)  $x-2>0 \text { and } x+3>0$

or

(ii)  $x-2<0 \text { and } x+3<0$

Case (i)       $\begin{aligned} & x-2>0 \text { and so } x>2 \\ & x+3>0 \text { and so } x>-3 \end{aligned}$

Both of these inequalities are satisfied only when x > 2.

Case (ii)            $\begin{aligned} & x-2<0 \text { and so } x<2 \\& x+3<0 \text { and so } x<-3 \end{aligned}$

Both of these inequalities are satisfied only when x < —3.
In summary, x² + x — 6 > 0 when either x > 2 or x < —3.

(b)   The quadratic expression x² + 8x + 1 does not factorize and so the technique of completing the square is used.

Hence

Using the result after Example 1.36 we may write