Quadratic Inequalities: Practice

Set up the inequality $h(t) \ge 48$ and move everything to one side.

$-16t^2 + 64t - 48 \ge 0 \;\Longleftrightarrow\; t^2 - 4t + 3 \le 0$

Dividing by $-16$ reversed the symbol. Factor as $(t - 1)(t - 3) \le 0$, with roots $1$ and $3$. The upward parabola is at or below zero between them, so the ball is at least $48$ feet high on $[1, 3]$.

Factor the non-monic quadratic.

$2x^2 - 5x - 3 = (2x + 1)(x - 3)$

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

Move everything to one side and recognize the perfect square.

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

A real square is never negative, so no $x$ can make it strictly less than zero. The solution set is the empty set $\varnothing$.

A single repeated root occurs exactly when the discriminant is zero.

$\Delta = (-8)^2 - 4(1)(k) = 64 - 4k = 0$

Solving gives $k = 16$. Then $x^2 - 8x + 16 = (x - 4)^2$, whose only root is the repeated value $4$.

Factor to find the roots.

$x^2 - 2x - 15 = (x - 5)(x + 3)$

The roots are $-3$ and $5$, and the upward parabola is positive outside them. The strict symbol excludes the roots, giving $(-\infty, -3) \cup (5, \infty)$.

An upward parabola stays at or above the axis exactly when it never dips below it, that is when the discriminant is at most zero.

$\Delta = k^2 - 16 \le 0 \;\Rightarrow\; -4 \le k \le 4$

At $k = \pm 4$ the parabola touches the axis, where $\ge 0$ still holds, so the endpoints are included. The answer is $[-4, 4]$.

Factor the non-monic quadratic.

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

The roots are $-1$ and $\tfrac{1}{3}$, and the upward parabola is negative between them. The strict symbol keeps the endpoints open, giving $\left(-1, \tfrac{1}{3}\right)$.

Move everything to one side.

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

The roots are $-2$ and $4$, and the upward parabola is at or above zero outside them. Including the endpoints for $\ge$, the answer is $(-\infty, -2] \cup [4, \infty)$.

The inequality has no solution exactly when the upward parabola never dips below the axis, that is when the discriminant is at most zero.

$\Delta = k^2 - 4 \le 0 \;\Rightarrow\; -2 \le k \le 2$

At $k = \pm 2$ the parabola touches the axis but still never goes below it, so $< 0$ remains unsolvable and the endpoints are included. The answer is $[-2, 2]$.

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

$-2x^2 + 5x + 3 \ge 0 \;\Longleftrightarrow\; 2x^2 - 5x - 3 \le 0$

Factor as $(2x + 1)(x - 3)$, with roots $-\tfrac{1}{2}$ and $3$. The upward parabola is at or below zero between them, so the answer is $\left[-\tfrac{1}{2}, 3\right]$.

A real square is never negative, so adding $1$ keeps the value positive for every input.

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

The expression is at or above zero for all real $x$ (in fact strictly positive), so the solution set is all of $\mathbb{R}$.

The quadratic is a perfect square.

$4x^2 - 12x + 9 = (2x - 3)^2$

A square is never negative, so $\le 0$ holds only where it is exactly zero, at $x = \tfrac{3}{2}$. The solution set is $\left\{\tfrac{3}{2}\right\}$.