Quadratic Inequalities: Additional Practice (Set 2)

The discriminant is $1 - 4 = -3 < 0$, so there are no real roots.

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

The upward parabola floats above the axis, so the inequality holds for all real numbers.

Multiply by $-1$ and flip the symbol.

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

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

The quadratic is a perfect square with a single (repeated) root.

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

A square is positive everywhere except where it is zero, so $(x - 4)^2 > 0$ holds for every $x$ except $x = 4$, giving $x < 4$ or $x > 4$.

Move every term to one side, then factor.

$x^2 + 3 \ge 4x \Rightarrow x^2 - 4x + 3 = (x - 1)(x - 3) \ge 0$

The upward parabola is positive outside the roots $1$ and $3$, and the inclusive symbol keeps them, so $x \le 1$ or $x \ge 3$.

Bring all terms to one side, then factor.

$2x^2 + 3 > 7x \Rightarrow 2x^2 - 7x + 3 = (2x - 1)(x - 3) > 0$

The roots are $\tfrac{1}{2}$ and $3$; the upward parabola is positive outside them, and the strict symbol excludes them, so $x < \tfrac{1}{2}$ or $x > 3$.

Move every term to one side, then factor.

$x^2 \le 2x + 8 \Rightarrow x^2 - 2x - 8 = (x - 4)(x + 2) \le 0$

The upward parabola is negative between the roots $-2$ and $4$, and the inclusive symbol keeps them, so $-2 \le x \le 4$.

Move everything to one side.

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

A square is never negative, so it can never be strictly less than $0$. There is no solution.

Factor and solve the inequality first.

$x^2 - 7x + 10 = (x - 2)(x - 5) \le 0 \Rightarrow 2 \le x \le 5$

The integers from $2$ to $5$ inclusive are $2, 3, 4, 5$, so there are $4$ integer solutions.

For an upward parabola to stay above the axis for every $x$, it must never touch the axis, so its discriminant must be negative.

$k^2 - 4(1)(9) < 0 \Rightarrow k^2 < 36 \Rightarrow -6 < k < 6$

At $k = \pm 6$ the parabola would touch the axis (equalling $0$ at its vertex), so the bound is strict.

The solution $-2 \le x \le 5$ is the band between the roots, so the roots are $-2$ and $5$. Use the sum and product of roots.

$b = -(-2 + 5) = -3, \qquad c = (-2)(5) = -10$

So $b + c = -3 + (-10) = -13$.

Multiply by $-1$ and flip, or read the downward parabola directly.

$3 - 2x - x^2 \ge 0 \Rightarrow x^2 + 2x - 3 = (x + 3)(x - 1) \le 0$

The upward parabola is negative between the roots $-3$ and $1$, and the inclusive symbol keeps them, so $-3 \le x \le 1$.

Factor and solve the inequality first.

$x^2 - 5x + 4 = (x - 1)(x - 4) < 0 \Rightarrow 1 < x < 4$

The integers strictly between $1$ and $4$ are $2$ and $3$, so there are $2$ integer solutions.