Quadratic Inequalities - Math 2

1. The quadratic function $f(x) = -2x^2 + 8x - 6$ is graphed on a coordinate plane. For which values of $x$ is $f(x) > 0$?
   Answer: $1 < x < 3$ To solve $f(x) > 0$, first find where $f(x) = 0$. Setting $-2x^2 + 8x - 6 = 0$, divide by -2: $x^2 - 4x + 3 = 0$. Factoring: $(x-1)(x-3) = 0$, so $x = 1$ or $x = 3$. Since the coefficient of $x^2$ is negative (-2), the parabola opens downward. Therefore, $f(x) > 0$ between the roots: $1 < x < 3$. Choice A gives the region where $f(x) < 0$. Choice C includes the boundary points where $f(x) = 0$. Choice D uses incorrect roots.
2. The quadratic inequality $ax^2 + bx + c \leq 0$ has solution set $[2, 5]$. If the same quadratic is written as $a(x - h)^2 + k \leq 0$, what are the values of $h$ and $k$?
   Answer: $h = 3.5, k = -2.25a$ where $a < 0$ Since the solution set is the closed interval $[2, 5]$, the quadratic must be negative or zero on this finite interval, which means the parabola opens downward, so $a < 0$. The vertex (where the maximum occurs) is at $h = \frac{2 + 5}{2} = 3.5$. Since the quadratic has roots at $x = 2$ and $x = 5$, we can write it as $a(x - 2)(x - 5)$. At the vertex $x = 3.5$: $k = a(3.5 - 2)(3.5 - 5) = a(1.5)(-1.5) = -2.25a$. Therefore $h = 3.5$ and $k = -2.25a$ where $a < 0$.
3. The solution set of $(x-1)^2(x+2) ≥ 0$ includes which of the following intervals?
   Answer: $[1, ∞)$ The expression $(x-1)^2(x+2)$ has zeros at $x = 1$ (with multiplicity 2) and $x = -2$ (with multiplicity 1). Since $(x-1)^2 ≥ 0$ for all $x$, the sign of the expression depends only on the sign of $(x+2)$. For $x < -2$: $(x+2) < 0$, so the expression is negative. For $x > -2$: $(x+2) > 0$, so the expression is non-negative. At $x = -2$: the expression equals 0. At $x = 1$: the expression equals 0. Therefore, $(x-1)^2(x+2) ≥ 0$ when $x ∈ [-2, ∞)$. This includes the interval $[1, ∞)$. Choice A represents where the expression is negative. Choice B represents part of where the expression is positive but excludes endpoints. Choice D is too broad as it should start at $-2$, not exclude it.
4. A quadratic inequality $ax^2 + bx + c ≤ 0$ has solution set $[-4, 2]$. If $a > 0$, what is the value of $\frac{b}{a}$?
   Answer: $2$ Since $a > 0$ and the solution set is $[-4, 2]$, the parabola opens upward and is non-positive between its roots at $x = -4$ and $x = 2$. The quadratic can be written as $a(x + 4)(x - 2) ≤ 0$. Expanding: $a(x^2 + 2x - 8) = ax^2 + 2ax - 8a$. Comparing with $ax^2 + bx + c$, we have $b = 2a$, so $\frac{b}{a} = 2$. Choice B gives the negative of the correct answer. Choice C represents $b$ when $a = 3$. Choice D represents $c$ when $a = 1$.
5. A student creates a sign chart for $\frac{(x + 2)^2(x - 1)}{x - 3} \leq 0$ and concludes the solution is $x \in (-\infty, 1] \cup (3, \infty)$. What error did the student most likely make?
   Answer: Forgot to include $x = -2$ in the final solution set The correct analysis: $(x + 2)^2 \geq 0$ always and equals 0 only at $x = -2$. The factor $(x - 1)$ changes from negative to positive at $x = 1$. The denominator $(x - 3)$ changes from negative to positive at $x = 3$ (excluded from domain). Sign analysis shows the expression is negative on $(1, 3)$ and zero at $x = -2$ and $x = 1$. The correct solution is $\{-2\} \cup [1, 3)$. The student's answer $(-\infty, 1] \cup (3, \infty)$ suggests they correctly found most intervals but forgot that $x = -2$ makes the entire expression equal to 0, satisfying the inequality.
6. Consider the quadratic inequality $2x^2 + kx - 3 > 0$. If this inequality has no solution, what is the range of possible values for $k$?
   Answer: No such values of $k$ exist For the quadratic $2x^2 + kx - 3$ with $a = 2 > 0$, the parabola opens upward. For the inequality $2x^2 + kx - 3 > 0$ to have no solution, the parabola would need to be entirely below the x-axis, meaning $2x^2 + kx - 3 ≤ 0$ for all $x$. However, since $a = 2 > 0$, the parabola opens upward and goes to $+∞$ as $x → ±∞$. Therefore, there will always be some values of $x$ for which $2x^2 + kx - 3 > 0$, regardless of the value of $k$. No such values of $k$ exist. Choice B gives the range where the quadratic has no real roots. Choice C gives the complement of that range. Choice A incorrectly suggests all values work.
7. Given that $f(x) = x^2 + px + q$ where $p$ and $q$ are constants, and the solution to $f(x) \leq 0$ is $[-3, 2]$, what is the solution to $f(x + 1) > 0$?
   Answer: $(-\infty, -4) \cup (1, \infty)$ Since $f(x) \leq 0$ has solution $[-3, 2]$, the quadratic $f(x)$ has zeros at $x = -3$ and $x = 2$, and opens upward. So $f(x) = a(x + 3)(x - 2)$ where $a > 0$. Thus $f(x) > 0$ when $x < -3$ or $x > 2$. For $f(x + 1) > 0$, we substitute $u = x + 1$, giving $f(u) > 0$ when $u < -3$ or $u > 2$. Converting back to $x$: $x + 1 < -3$ gives $x < -4$, and $x + 1 > 2$ gives $x > 1$. Therefore, $f(x + 1) > 0$ when $x \in (-\infty, -4) \cup (1, \infty)$. The horizontal shift left by 1 unit moves the solution intervals accordingly.
8. If $p(x) = x^2 + mx + n$ and the solution set to $p(x) < 0$ is $(-4, 1)$, what is the solution set to $p(x - 3) \geq 0$?
   Answer: $(-\infty, 1] \cup [7, \infty)$ Since $p(x) < 0$ on $(-4, 1)$, the quadratic $p(x)$ has zeros at $x = -4$ and $x = 1$, and opens upward (positive leading coefficient). Thus $p(x) \geq 0$ when $x \leq -4$ or $x \geq 1$. For $p(x - 3) \geq 0$, we substitute $u = x - 3$, giving $p(u) \geq 0$ when $u \leq -4$ or $u \geq 1$. Converting back: $x - 3 \leq -4$ gives $x \leq -1$, and $x - 3 \geq 1$ gives $x \geq 4$. The horizontal shift moves the solution set 3 units right.
9. A quadratic function $f(x) = ax^2 + bx + c$ satisfies $f(x) < 0$ for $x ∈ (-2, 5)$ and $f(x) > 0$ for $x ∈ (-∞, -2) ∪ (5, ∞)$. If $f(0) = -10$, what is the value of $a$?
   Answer: $a = 1$ Since $f(x) > 0$ for $x ∈ (-∞, -2) ∪ (5, ∞)$ and $f(x) < 0$ for $x ∈ (-2, 5)$, the parabola opens upward ($a > 0$) with roots at $x = -2$ and $x = 5$. Therefore, $f(x) = a(x + 2)(x - 5) = a(x^2 - 3x - 10)$. Given $f(0) = -10$: $f(0) = a(0 + 2)(0 - 5) = a(2)(-5) = -10a = -10$. Solving: $a = 1$. Choice B would give $f(0) = -20$. Choice C would make the parabola open downward, contradicting the given information. Choice D would give $f(0) = -5$.
10. The inequality $|x^2 - 4x| ≤ 5$ is equivalent to which of the following compound inequalities?
    Answer: $-1 ≤ x ≤ 5$ The inequality $|x^2 - 4x| ≤ 5$ is equivalent to $-5 ≤ x^2 - 4x ≤ 5$. This gives us two inequalities: (1) $x^2 - 4x ≥ -5$, which becomes $x^2 - 4x + 5 ≥ 0$. The discriminant is $16 - 20 = -4 < 0$, so this is always true. (2) $x^2 - 4x ≤ 5$, which becomes $x^2 - 4x - 5 ≤ 0$. Factoring: $(x - 5)(x + 1) ≤ 0$. This gives $-1 ≤ x ≤ 5$. Since the first condition is always satisfied, the solution is $-1 ≤ x ≤ 5$. Choice B incorrectly splits the interval. Choice C gives the complement of the correct answer. Choice D represents the interval where $x^2 - 4x ≤ 0$.
11. Consider the system of inequalities: $x^2 - 5x + 6 < 0$ and $x^2 - x - 12 \geq 0$. What is the solution set for this system?
    Answer: $\emptyset$ (no solution) For the first inequality: $x^2 - 5x + 6 = (x - 2)(x - 3) < 0$. This quadratic is negative between its roots, so the solution is $2 < x < 3$. For the second inequality: $x^2 - x - 12 = (x - 4)(x + 3) \geq 0$. This quadratic is non-negative outside its roots, so the solution is $x \leq -3$ or $x \geq 4$. The system requires both conditions to hold simultaneously: $x \in (2, 3)$ AND $x \in (-\infty, -3] \cup [4, \infty)$. Since the interval $(2, 3)$ does not intersect with $(-\infty, -3] \cup [4, \infty)$, there is no solution.
12. The inequality $x^2 - kx + k ≥ 0$ is satisfied by all real numbers. Which of the following describes the possible values of $k$?
    Answer: $0 ≤ k ≤ 4$ For the quadratic $x^2 - kx + k$ to be non-negative for all real numbers, its discriminant must be non-positive (so it has at most one real root and opens upward). The discriminant is $(-k)^2 - 4(1)(k) = k^2 - 4k = k(k-4)$. For $k(k-4) ≤ 0$, we need $k$ and $(k-4)$ to have opposite signs or one to be zero. This occurs when $0 ≤ k ≤ 4$. Choice A gives the complement condition. Choice C excludes the boundary cases. Choice D only includes the boundary points where the discriminant equals zero.
13. The solution to the inequality $x^2 - 4x + k > 0$ is all real numbers except for a single point. What is the value of $k$, and what is that exceptional point?
    Answer: $k = 4$ and the point is $x = 2$ For a quadratic $ax^2 + bx + c > 0$ to be positive for all real numbers except one point, the discriminant must equal zero (perfect square) and $a > 0$. Here, $\Delta = 16 - 4k = 0$, so $k = 4$. The quadratic becomes $x^2 - 4x + 4 = (x - 2)^2$. Since $(x - 2)^2 \geq 0$ for all real $x$ and equals zero only when $x = 2$, the inequality $(x - 2)^2 > 0$ holds for all real numbers except $x = 2$.

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