Quadratic Inequalities: Additional Practice (Set 1)

Factor to find the roots.

$x^2 - x - 20 = (x - 5)(x + 4)$

The roots are $-4$ and $5$, and the upward parabola is at or above zero outside them. Including the endpoints for $\ge$, the answer is $(-\infty, -4] \cup [5, \infty)$.

Factor to find the roots.

$x^2 + 3x - 10 = (x + 5)(x - 2)$

The roots are $-5$ and $2$, and the upward parabola is negative between them. The strict symbol keeps the endpoints open, giving $(-5, 2)$.

An upward parabola is strictly positive everywhere exactly when its discriminant is negative.

$\Delta = 16 - 4k < 0 \;\Rightarrow\; k > 4$

At $k = 4$ the parabola touches the axis, where $> 0$ fails, so the endpoint is excluded. The answer is $(4, \infty)$.

Factor the difference of squares.

$4x^2 - 1 = (2x - 1)(2x + 1)$

The roots are $-\tfrac{1}{2}$ and $\tfrac{1}{2}$, and the upward parabola is at or below zero between them. Including the endpoints for $\le$, the answer is $\left[-\tfrac{1}{2}, \tfrac{1}{2}\right]$.

The quadratic is a perfect square.

$x^2 - 10x + 25 = (x - 5)^2$

A square is zero only at $x = 5$ and positive everywhere else. The strict symbol $> 0$ removes that single point, so the answer is $(-\infty, 5) \cup (5, \infty)$.

A repeated root occurs exactly when the discriminant is zero.

$\Delta = k^2 - 4(1)(25) = k^2 - 100 = 0$

Solving gives $k^2 = 100$, so $k = 10$ or $k = -10$. Both values produce a perfect square, $(x \pm 5)^2$.

An upward parabola stays at or above the axis exactly when it never dips strictly below it.

$f(x) \ge 0 \text{ for all } x \;\Longleftrightarrow\; \Delta \le 0$

If $\Delta < 0$ the parabola floats entirely above the axis, and if $\Delta = 0$ it touches the axis at one point where $f = 0$, which still satisfies $\ge 0$. Both are allowed, so the condition is $\Delta \le 0$, not one of them alone.

Factor the non-monic quadratic.

$2x^2 + 3x - 2 = (2x - 1)(x + 2)$

The roots are $-2$ and $\tfrac{1}{2}$, and the upward parabola is positive outside them. The strict symbol excludes the roots, giving $(-\infty, -2) \cup \left(\tfrac{1}{2}, \infty\right)$.

Move everything to one side.

$x^2 - 5x + 6 \le 0 \;\Rightarrow\; (x - 2)(x - 3) \le 0$

The roots are $2$ and $3$, and the upward parabola is at or below zero between them. Including the endpoints for $\le$, the answer is $[2, 3]$.

With $a < 0$ the parabola opens downward, so it is positive between its roots rather than outside them.

$a < 0 \;\Rightarrow\; f > 0 \text{ on } 1 < x < 4$

The strict symbol excludes the roots, giving the open interval $(1, 4)$.

Factor the non-monic quadratic.

$6x^2 - x - 1 = (3x + 1)(2x - 1)$

The roots are $-\tfrac{1}{3}$ and $\tfrac{1}{2}$, and the upward parabola is at or below zero between them. Including the endpoints for $\le$, the answer is $\left[-\tfrac{1}{3}, \tfrac{1}{2}\right]$.

Expand and move the constant across so one side is zero.

$x^2 - 6x + 8 < 0 \;\Rightarrow\; (x - 2)(x - 4) < 0$

The roots are $2$ and $4$, and the upward parabola is negative between them. The strict symbol keeps the endpoints open, giving $(2, 4)$.