Further Maths

When are these quadratic inequalities true together?

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When are these quadratic inequalities true together?

Question

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

$x < -2$ ;
$-1 < x < 1$ ;
$x > -1$ ;
$x > -2$ .

Suggestion

Can we write the quadratic expressions in a more useful form?

What would the inequalities look like on a graph?

Solution

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.

The two graphs x squared + 3 x + 2 and x squared + x minus 2. The solution first inequality is the is the part of the x axis above the first curve, and the second is the part below the second curve, as previously calculated.

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.