No Real Intercepts & Complex Roots - Math 2

Consider the family of quadratics $f_t(x) = x^2 + 2tx + (t^2 + 4)$ where $t$ is a real parameter. For which values of $t$ does $f_t(x)$ have no real x-intercepts, and what can be said about the complex roots?
Answer: For all real $t$ , there are no real x-intercepts, and the complex roots are always $-t \pm 2i$ The discriminant is $(2t)^2 - 4(1)(t^2 + 4) = 4t^2 - 4t^2 - 16 = -16 < 0$ for all real t. So there are never real x-intercepts. Completing the square: $f_t(x) = (x + t)^2 + 4$ . Setting equal to zero: $(x + t)^2 = -4$ , so $x + t = \pm 2i$ , giving $x = -t \pm 2i$ . The roots are always $-t \pm 2i$ regardless of t. Choice B incorrectly restricts t values. Choice C incorrectly suggests the condition is |t| < 2 and that imaginary parts vary with t (they're always ±2). Choice D incorrectly suggests there are real roots for some t values.
The quadratic function $h(x) = 2x^2 - 8x + m$ has no real x-intercepts. If $m$ is the smallest integer value that satisfies this condition, what is the sum of the imaginary parts of the two complex roots?
Answer: $0$ For no real x-intercepts, the discriminant must be negative: $(-8)^2 - 4(2)(m) < 0$ , so $64 - 8m < 0$ , giving $m > 8$ . The smallest integer is $m = 9$ . The roots are $\frac{8 \pm \sqrt{64-72}}{4} = \frac{8 \pm \sqrt{-8}}{4} = \frac{8 \pm 2i\sqrt{2}}{4} = 2 \pm \frac{i\sqrt{2}}{2}$ . The imaginary parts are $+\frac{\sqrt{2}}{2}$ and $-\frac{\sqrt{2}}{2}$ , which sum to 0. This is always true for complex conjugate pairs. Choice B and D suggest non-zero sums, which is impossible for conjugate pairs. Choice C gives a specific non-zero value, also impossible.
A parabola has equation $y = ax^2 + bx + c$ where $a, b, c$ are real and $a \neq 0$ . The parabola intersects the line $y = d$ at two complex points. What does this tell us about the relationship between $d$ and the parabola?
Answer: The value $d$ is less than the minimum value of the parabola when $a > 0$ , or greater than the maximum value when $a < 0$ When you encounter questions about parabolas intersecting horizontal lines at complex points, you're dealing with the relationship between a parabola's range and where horizontal lines can intersect it in the complex plane. To find where $y = ax^2 + bx + c$ intersects $y = d$ , you solve $ax^2 + bx + c = d$ , or $ax^2 + bx + (c-d) = 0$ . The discriminant is $b^2 - 4a(c-d)$ . For complex solutions, this discriminant must be negative: $b^2 - 4a(c-d) < 0$ . The vertex of the parabola occurs at $x = -\frac{b}{2a}$ with y-coordinate $y_v = c - \frac{b^2}{4a}$ . When $a > 0$ , the parabola opens upward, so $y_v$ is the minimum value. When $a < 0$ , it opens downward, so $y_v$ is the maximum value. From our discriminant condition: $b^2 < 4a(c-d)$ , which gives us $\frac{b^2}{4a} > c-d$ , so $d > c - \frac{b^2}{4a} = y_v$ when $a > 0$ , and $d < y_v$ when $a < 0$ . Wait—this means the line must be beyond the vertex to have no real intersections. Actually, for complex intersections, $d$ must be outside the parabola's range entirely. This means $d$ is less than the minimum when $a > 0$ , or greater than the maximum when $a < 0$ . Choice A is wrong because not all horizontal lines give complex intersections. Choice B describes tangency (one real solution). Choice C describes the case for two real solutions. Choice D correctly identifies when intersections are purely complex. Remember: complex intersections occur when the horizontal line completely "misses" the parabola's actual range.
The quadratic functions $f(x) = 2x^2 + 8x + 10$ and $g(x) = x^2 + 4x + 5$ are related by $f(x) = 2g(x)$ . How do their complex roots compare?
Answer: They have the same complex roots because vertical stretching doesn't affect x-intercepts Since f(x) = 2g(x), the equation f(x) = 0 becomes 2g(x) = 0, which simplifies to g(x) = 0. Therefore, f(x) and g(x) have identical roots. We can verify: g(x) = (x+2)² + 1, so roots are -2 ± i. For f(x) = 2((x+2)² + 1) = 2(x+2)² + 2, setting equal to zero gives (x+2)² + 1 = 0, yielding the same roots -2 ± i. Choice A is wrong because it suggests they're the same 'since' it's a stretch, when actually the roots are the same despite the stretch having no effect on roots. Choice C incorrectly applies scaling to the roots themselves. Choice D confuses scaling the function with scaling the roots.
A quadratic function $f(x) = x^2 + px + q$ satisfies $f(2) = f(6) = 0$ . If the related equation $x^2 + px + (q + k) = 0$ has no real solutions, which inequality must $k$ satisfy?
Answer: $k > 4$ Since $f(x) = x^2 + px + q$ has zeros at $x = 2$ and $x = 6$ , we can write $f(x) = (x-2)(x-6) = x^2 - 8x + 12$ . Therefore $p = -8$ and $q = 12$ . The related equation becomes $x^2 - 8x + (12 + k) = 0$ . For this to have no real solutions, the discriminant must be negative: $\Delta = (-8)^2 - 4(1)(12 + k) = 64 - 4(12 + k) = 64 - 48 - 4k = 16 - 4k < 0$ . Solving: $16 - 4k < 0 \Rightarrow 16 < 4k \Rightarrow k > 4$ . Choice A would allow some values where real solutions exist. Choice B is too weak since $k = 1$ would give $\Delta = 12 > 0$ . Choice D is too restrictive since $k = 5$ already ensures no real solutions.
The complex roots of $f(x) = 2x^2 - 12x + 25$ are $r_1$ and $r_2$ . If $g(x) = f(x + 3)$ , how do the complex roots of $g(x) = 0$ relate to those of $f(x) = 0$ ?
Answer: The roots of $g(x) = 0$ are $r_1 - 3$ and $r_2 - 3$ If g(x) = f(x + 3), then g(x) = 0 means f(x + 3) = 0. Let u = x + 3, so f(u) = 0, which means u = r₁ or u = r₂. Since u = x + 3, we have x + 3 = r₁ or x + 3 = r₂, giving x = r₁ - 3 or x = r₂ - 3. We can verify: f(x) = 2(x - 3)² + 7 has roots 3 ± i√(7/2). Then g(x) = f(x + 3) = 2((x + 3) - 3)² + 7 = 2x² + 7, with roots ±i√(7/2). Indeed, these are 3 ± i√(7/2) - 3 = ±i√(7/2). Choice A gets the sign wrong. Choice C is incorrect because horizontal shifts do affect root locations. Choice D is correct about imaginary parts but incomplete.
Two quadratic functions $f(x) = x^2 + 4x + 7$ and $g(x) = -x^2 + 2x - 5$ both have no real x-intercepts. Which statement correctly compares their complex roots?
Answer: The function $g(x)$ has complex roots with larger imaginary parts because its discriminant is more negative For f(x): discriminant = 16 - 28 = -12, vertex at x = -2. For g(x): discriminant = 4 - 20 = -16, vertex at x = 1. Since g(x) has a more negative discriminant (-16 vs -12), its complex roots have larger imaginary parts. The roots are f(x): $-2 \pm i\sqrt{3}$ and g(x): $1 \pm 2i$ . Choice A is wrong because the vertices have different x-coordinates (-2 vs 1). Choice B reverses the comparison. Choice D is wrong because the discriminants are different (-12 vs -16).
A parabola opens upward and has its vertex at $(3, 5)$ . If this parabola has no real x-intercepts, which of the following could be the complex roots of the corresponding quadratic equation?
Answer: $3 \pm 2i$ Since the vertex is at (3, 5) and the parabola opens upward with no real x-intercepts, the parabola never touches the x-axis. The quadratic can be written as $a(x-3)^2 + 5 = 0$ where $a > 0$ . This gives $(x-3)^2 = -\frac{5}{a}$ , so $x-3 = \pm i\sqrt{\frac{5}{a}}$ , meaning $x = 3 \pm i\sqrt{\frac{5}{a}}$ . The real part of the complex roots equals the x-coordinate of the vertex (3). Choice B has the wrong real part (-3). Choice C uses the y-coordinate as the real part. Choice D has the correct real part but would require a discriminant that doesn't match the vertex form.
Two parabolas $P_1: y = x^2 + 2x + 5$ and $P_2: y = x^2 + 2x + 2$ have no real x-intercepts. If $P_2$ is shifted vertically to create a parabola $P_3$ that has exactly one real x-intercept, what happens to the relationship between the complex roots of $P_1$ and $P_3$ ?
Answer: The complex roots of $P_1$ have non-zero imaginary parts while $P_3$ has a double root with zero imaginary part P₁: y = (x+1)² + 4 has roots -1 ± 2i. P₂: y = (x+1)² + 1 has roots -1 ± i. For P₃ to have exactly one real x-intercept, P₂ must be shifted down by 1 unit, giving P₃: y = (x+1)² with double root x = -1. So P₁ has complex roots -1 ± 2i (imaginary parts ±2), while P₃ has repeated real root -1 (imaginary part 0). Choice A is wrong because having the same vertex x-coordinate doesn't make the roots identical. Choice B is misleading because P₃ does have a repeated root, which could be considered as having complex roots with zero imaginary part. Choice C is correct about the nature but doesn't emphasize the imaginary part distinction.
A quadratic function $f(x)$ has complex zeros $3 + 2i$ and $3 - 2i$ . If $f(0) = 26$ , which statement about the graph of $y = f(x)$ is correct?
Answer: The graph intersects the y-axis above the x-axis and has a minimum point above the x-axis Since the zeros are $3 \pm 2i$ , the quadratic can be written as $f(x) = a(x - (3+2i))(x - (3-2i)) = a((x-3)^2 + 4) = a(x^2 - 6x + 13)$ . Using $f(0) = 26$ : $f(0) = a(0 - 0 + 13) = 13a = 26$ , so $a = 2$ . Therefore $f(x) = 2(x^2 - 6x + 13) = 2x^2 - 12x + 26$ . Since $a = 2 > 0$ , the parabola opens upward and has a minimum. The vertex is at $x = \frac{12}{4} = 3$ , and $f(3) = 2(9) - 12(3) + 26 = 18 - 36 + 26 = 8 > 0$ . The y-intercept is $f(0) = 26 > 0$ , so the graph intersects the y-axis above the x-axis. Since the vertex is at $(3, 8)$ with $8 > 0$ , the minimum point is above the x-axis. Choice A is wrong because $f(0) = 26 > 0$ and the vertex is not at $(3,0)$ . Choice C is wrong because the parabola has a minimum, not maximum. Choice D is wrong because $f(0) > 0$ and it's a minimum, not maximum.
Two quadratic functions $p(x) = x^2 + bx + c$ and $q(x) = -x^2 + bx + c$ share the same values for $b$ and $c$ . If $p(x)$ has no real zeros, what can be concluded about the zeros of $q(x)$ ?
Answer: $q(x)$ must have two distinct real zeros since the discriminant changes when the leading coefficient changes sign For $p(x) = x^2 + bx + c$ , the discriminant is $\Delta_p = b^2 - 4c$ . For $q(x) = -x^2 + bx + c$ , the discriminant is $\Delta_q = b^2 - 4(-1)(c) = b^2 + 4c$ . Since $p(x)$ has no real zeros, $\Delta_p = b^2 - 4c < 0$ , which means $b^2 < 4c$ . For $q(x)$ : $\Delta_q = b^2 + 4c > 0$ since both $b^2 \geq 0$ and $4c > b^2 \geq 0$ . Therefore $q(x)$ must have two distinct real zeros. The key insight is that changing the sign of the leading coefficient changes the discriminant formula.
Consider the quadratic function $h(x) = -x^2 + 4x - 7$ . Based on the graph of this function, which statement correctly describes the relationship between its complex roots and the graph's position?
Answer: The graph has a maximum point below the x-axis, and the complex roots have real part equal to the x-coordinate of the vertex For $h(x) = -x^2 + 4x - 7$ , the vertex occurs at $x = -\frac{4}{2(-1)} = 2$ . The vertex is at $(2, h(2)) = (2, -4 + 8 - 7) = (2, -3)$ . Since $a = -1 < 0$ , this parabola opens downward with a maximum at $(2, -3)$ . The discriminant is $\Delta = 16 - 4(-1)(-7) = 16 - 28 = -12 < 0$ , confirming complex roots. The complex roots are $x = \frac{4 \pm \sqrt{-12}}{2} = \frac{4 \pm 2\sqrt{3}i}{2} = 2 \pm \sqrt{3}i$ . The real part of both roots is 2, which equals the x-coordinate of the vertex. Choice A is wrong because the sign of the real parts isn't determined by the vertex position. Choice B is wrong because the maximum is below the x-axis. Choice D is wrong because this parabola has a maximum, not minimum.
A student claims that if a quadratic function $f(x) = ax^2 + bx + c$ has complex roots $r_1 = p + qi$ and $r_2 = p - qi$ where $q \neq 0$ , then the y-intercept of the graph must equal $p^2 + q^2$ . Is this claim correct?
Answer: No, because the y-intercept is $c = a(p^2 + q^2)$ , which equals $p^2 + q^2$ only when $a = 1$ The y-intercept is c. Using Vieta's formulas and the factored form: if the roots are $p \pm qi$ , then $f(x) = a(x-(p+qi))(x-(p-qi)) = a((x-p)^2 + q^2) = a(x^2 - 2px + p^2 + q^2)$ . So $c = a(p^2 + q^2)$ . The student's claim is only true when $a = 1$ . Choice A incorrectly describes what $p^2 + q^2$ represents geometrically. Choice B is partially correct but doesn't address that the claim is generally false. Choice D is wrong because there is a relationship, just not the one the student claimed.

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