Quadratic Inequalities: Practice

The quadratic does not factor nicely, so use the quadratic formula with $a = 1$, $b = -4$, $c = 1$.

$x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$

The upward parabola is negative between the roots, and the inclusive $\le$ keeps the endpoints, so $2 - \sqrt{3} \le x \le 2 + \sqrt{3}$.

Use the quadratic formula with $a = 1$, $b = -6$, $c = 7$.

$x = \frac{6 \pm \sqrt{36 - 28}}{2} = 3 \pm \sqrt{2}$

The upward parabola is positive outside the roots, and the strict symbol excludes them, so $x < 3 - \sqrt{2}$ or $x > 3 + \sqrt{2}$.

Use the quadratic formula with $a = 1$, $b = 2$, $c = -2$.

$x = \frac{-2 \pm \sqrt{4 + 8}}{2} = -1 \pm \sqrt{3}$

The upward parabola is positive outside the roots, and the inclusive $\ge$ keeps them, so $x \le -1 - \sqrt{3}$ or $x \ge -1 + \sqrt{3}$.

The discriminant is $0 - 4(1)(1) = -4 < 0$, so there are no real roots and the upward parabola never touches the axis.

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

The quadratic is always positive, so $x^2 + 1 > 0$ holds for all real numbers.

The discriminant is $0 - 4(1)(4) = -16 < 0$, so the upward parabola sits entirely above the axis.

$x^2 + 4 \ge 4 > 0 \text{ for every } x$

The quadratic is never negative, so $x^2 + 4 < 0$ has no solution.

Check the discriminant with $a = 1$, $b = -2$, $c = 5$.

$b^2 - 4ac = 4 - 20 = -16 < 0$

The upward parabola never reaches the axis, so it is always positive and never $\le 0$. There is no solution.

The quadratic is a perfect square.

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

A square is never negative, so $(x - 3)^2 \le 0$ can only hold where it equals $0$, at the single point $x = 3$.

The quadratic is a perfect square.

$x^2 + 4x + 4 = (x + 2)^2 \ge 0 \text{ for every } x$

A square is never negative, so $(x + 2)^2 \ge 0$ is true for all real numbers.

Move every term to one side, then factor.

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

The upward parabola is negative between the roots, and the inclusive symbol keeps them, so $2 \le x \le 3$.

Move everything to one side.

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

A square is never negative, so this holds for all real numbers.

Use the quadratic formula with $a = 1$, $b = -2$, $c = -4$.

$x = \frac{2 \pm \sqrt{4 + 16}}{2} = 1 \pm \sqrt{5}$

The upward parabola is negative between the roots, and the strict symbol excludes them, so $1 - \sqrt{5} < x < 1 + \sqrt{5}$.

The roots come from setting the quadratic to $0$.

$x^2 - 5 = 0 \Rightarrow x^2 = 5 \Rightarrow x = \pm\sqrt{5}$

The upward parabola is positive outside the roots, so $x < -\sqrt{5}$ or $x > \sqrt{5}$.