Quadratic Inequalities: Additional Practice (Set 2)

Two distinct real roots require a positive discriminant.

$\Delta = k^2 - 4(1)(9) = k^2 - 36 > 0$

This gives $k^2 > 36$, so $k < -6$ or $k > 6$. The set of such $k$ is $(-\infty, -6) \cup (6, \infty)$.

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

$-x^2 + 6x - 5 \ge 0 \;\Longleftrightarrow\; x^2 - 6x + 5 \le 0$

Factor as $(x - 1)(x - 5)$, with roots $1$ and $5$. The upward parabola is at or below zero between them, so the answer is $[1, 5]$.

An upward parabola stays strictly positive everywhere exactly when it never reaches the axis, that is when the discriminant is negative.

$\Delta = 36 - 4k < 0 \;\Rightarrow\; k > 9$

At $k = 9$ the discriminant is zero and the parabola touches the axis, so $> 0$ would fail there. The strict requirement gives $(9, \infty)$.

Move everything to one side and recognize the perfect square.

$x^2 - 10x + 25 < 0 \;\Rightarrow\; (x - 5)^2 < 0$

A real square is never negative, so nothing satisfies the strict inequality. The solution set is the empty set $\varnothing$.

The quadratic is a perfect square.

$9x^2 - 6x + 1 = (3x - 1)^2$

This is zero only at $x = \tfrac{1}{3}$ and positive everywhere else. The strict symbol $> 0$ removes that one point, so the solution is $\left(-\infty, \tfrac{1}{3}\right) \cup \left(\tfrac{1}{3}, \infty\right)$.

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

$\Delta = 36 - 4k \le 0 \;\Rightarrow\; k \ge 9$

At $k = 9$ the parabola touches the axis but does not cross below it, so $< 0$ is still unsolvable and the endpoint is included. The answer is $[9, \infty)$.

The closed interval $[-1, 4]$ tells you the roots are $-1$ and $4$, since those are where $f$ equals zero and the sign switches.

$f(x) \le 0 \text{ on } [-1, 4] \;\Rightarrow\; f(x) > 0 \text{ outside the roots}$

Where $f$ was at or below zero, its strict positive part is the complementary open rays $(-\infty, -1) \cup (4, \infty)$. The roots themselves belong to neither, because there $f = 0$.

Check the discriminant, then complete the square.

$\Delta = 2^2 - 4(1)(5) = -16 < 0, \qquad x^2 + 2x + 5 = (x + 1)^2 + 4$

The square is never negative, so the expression is at least $4 > 0$ for every $x$. The solution set is all of $\mathbb{R}$.

Completing the square shows the expression is always positive.

$x^2 + 2x + 5 = (x + 1)^2 + 4 \ge 4 > 0$

Since the value is never at or below zero, no $x$ satisfies $\le 0$, and the solution set is the empty set $\varnothing$.

Divide by the positive number $3$, keeping the symbol, then factor.

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

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

The two closed rays meet the roots at $-3$ and $5$, which are where the upward parabola equals zero.

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

Matching to $x^2 + bx + c$ gives $b = -2$ and $c = -15$.

Factor out $-3$ to reveal a perfect square.

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

Since $(x - 2)^2 \ge 0$, multiplying by $-3$ makes $-3(x - 2)^2 \le 0$ for every $x$, equal to zero only at $x = 2$. So $\ge 0$ holds at that single point, and the solution set is $\{2\}$.