Review question

Ref: R9989

## Solution

The inequalities $x^2 + 3x + 2 > 0$ and $x^2 + x < 2$ are met by all $x$ in the region:

1. $x < -2$;

2. $-1 < x < 1$;

3. $x > -1$;

4. $x > -2$.

What happens if we sketch the graphs of $y = x^2+3x+2$ and $y = x^2 + x -2$?

Note that $x^2 + 3x + 2 = (x+1)(x+2),$ so $y= 0$ for this curve when $x = -1$ or $x=-2$.

Similarly, $x^2 + x - 2 = (x-1)(x+2)$ so $y = 0$ for this curve when $x = 1$ or $x=-2$.

If we draw a sketch, we can quickly see the regions where each inequality is satisfied.

We need the region where the blue curve is below the $x$-axis, and where the red curve is above the $x$-axis.

We can see that both inequalities are satisfied in the range $-1 < x < 1$, and the answer is (b).

For this multiple choice question, we could have been even more efficient. The second inequality $x^2+x<2$ is only met by a finite interval of $x$ values as the quadratic tends to infinity for large values of $x$ (either positive or negative). The only option which offers a finite interval of $x$ values is (b), so this must be the correct answer.