Quadratic Inequalities: Practice

Write the factor as $x - (-2)$, so its root is $-2$. A linear factor is negative to the left of its root.

$x + 2 < 0 \;\Longleftrightarrow\; x < -2$

So $x + 2$ is negative for every $x$ less than $-2$.

With $a > 0$ the parabola opens upward, so it is positive outside its roots and negative between them.

$\text{positive when } x < -3 \text{ or } x > 1$

Written as a set, that is the union of two rays $(-\infty, -3) \cup (1, \infty)$.

Substitute each candidate and keep the one that makes the statement true.

$0^2 - 4 = -4 < 0 \text{ is true}$

The others give $3^2 - 4 = 5$, $(-5)^2 - 4 = 21$, and $2^2 - 4 = 0$, none of which is less than $0$. Only $x = 0$ works.

Set each factor equal to zero and solve.

$x - 3 = 0 \Rightarrow x = 3, \qquad x + 2 = 0 \Rightarrow x = -2$

So the roots are $3$ and $-2$. Notice the sign of each root is opposite to the sign shown in its factor.

The inequality is strict, so $4$ itself is not included, and the values run upward without bound.

$x > 4 \;\Longleftrightarrow\; (4, \infty)$

A strict inequality uses a parenthesis at the endpoint, and $\infty$ always takes a parenthesis.

The roots are $2$ and $5$, and the parabola opens upward, so it is positive outside the roots and negative between them.

$\text{positive when } x < 2 \text{ or } x > 5$

The symbol is strict, so the roots are excluded, giving $(-\infty, 2) \cup (5, \infty)$.

The roots are $2$ and $5$, and an upward parabola is negative exactly between its roots.

$\text{negative when } 2 < x < 5$

The symbol is strict, so the endpoints are open, giving the single interval $(2, 5)$.

The roots are $-1$ and $3$, and an upward parabola is at or below zero between its roots.

$\text{at most zero when } -1 \le x \le 3$

The symbol $\le$ includes the roots, so the endpoints are closed, giving $[-1, 3]$.

A zero discriminant means the two roots have merged into a single repeated root.

$\Delta = 0 \;\Rightarrow\; \text{one repeated real root}$

The parabola just touches the $x$-axis at its vertex instead of crossing it.

Read the sign of the leading coefficient $a$.

$a = -2 < 0 \;\Rightarrow\; \text{opens downward}$

A negative leading coefficient opens the parabola downward, with a highest point at the vertex.

The curve meets the axis where $y = 0$, which needs the square to be zero.

$(x - 4)^2 = 0 \;\Rightarrow\; x = 4$

A perfect square is zero only at its single repeated root, so the parabola touches the axis at $x = 4$.

Substitute $x = 1$ into the product and compare with the symbol.

$(1 - 1)(1 + 2) = 0 \cdot 3 = 0$

The result is $0$, and the symbol $\ge$ accepts zero, so $x = 1$ is a solution. Only a strict $>$ would reject it.