Quadratic Inequalities: Practice

Factor and find the roots.

$x^2 - 5x + 6 = (x - 2)(x - 3)$

The roots are $2$ and $3$, and the upward parabola is negative between them. The symbol is strict, so the endpoints are open, giving $(2, 3)$.

Factor the difference of squares.

$x^2 - 9 = (x - 3)(x + 3)$

The roots are $-3$ and $3$, and the upward parabola is positive outside them. The strict symbol excludes the roots, giving $(-\infty, -3) \cup (3, \infty)$.

Factor to find the roots.

$x^2 + 2x - 8 = (x + 4)(x - 2)$

The roots are $-4$ and $2$, and an upward parabola is at or below zero between them. Including the endpoints for $\le$, the answer is $[-4, 2]$.

Factor out the common $x$.

$x^2 - 7x = x(x - 7)$

The roots are $0$ and $7$, and the upward parabola is positive outside them. The strict symbol excludes the roots, giving $(-\infty, 0) \cup (7, \infty)$.

Move everything to one side first.

$x^2 - 16 \le 0 \;\Rightarrow\; (x - 4)(x + 4) \le 0$

The roots are $-4$ and $4$, and the upward parabola is at or below zero between them. With $\le$ the endpoints are closed, so the answer is $[-4, 4]$.

The leading coefficient is negative, so multiply by $-1$ and reverse the symbol.

$-x^2 + 4 > 0 \;\Longleftrightarrow\; x^2 - 4 < 0$

Now the roots are $-2$ and $2$, and the upward parabola is negative between them. The strict symbol gives the open interval $(-2, 2)$.

Make the leading coefficient positive by multiplying by $-1$ and reversing the symbol.

$-x^2 + x + 6 \ge 0 \;\Longleftrightarrow\; x^2 - x - 6 \le 0$

Factor as $(x - 3)(x + 2)$, with roots $-2$ and $3$. The upward parabola is at or below zero between them, and $\le$ includes the endpoints, so the answer is $[-2, 3]$.

Divide by the positive number $2$, which keeps the symbol pointing the same way.

$2x^2 - 8 < 0 \;\Longleftrightarrow\; x^2 - 4 < 0$

The roots are $-2$ and $2$, and the upward parabola is negative between them. The strict symbol gives $(-2, 2)$.

Recognize the perfect square.

$x^2 - 6x + 9 = (x - 3)^2$

A square is zero only at $x = 3$ and positive everywhere else. The strict symbol $> 0$ excludes that single point, leaving every other real number, so the answer is $(-\infty, 3) \cup (3, \infty)$.

The quadratic is a perfect square.

$x^2 - 6x + 9 = (x - 3)^2 \ge 0$

A square is never negative, so $\le 0$ can hold only where it equals zero, which is the single point $x = 3$. The solution set is $\{3\}$.

A real square is never negative, so add $1$ and the value stays positive.

$x^2 + 1 \ge 1 > 0 \text{ for every } x$

The expression is positive for all inputs (its discriminant is $-4 < 0$), so the solution set is all of $\mathbb{R}$.

A real square is never negative, so it is always at or above zero.

$(x - 1)^2 \ge 0 \text{ for every } x$

The inequality holds for all inputs, including $x = 1$ where the square is exactly zero, so the solution set is all of $\mathbb{R}$.