Math 2 Quiz: No Real Intercepts And Complex Roots

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No Real Intercepts And Complex RootsQuestion 1 of 13

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 $P_3$ ?

P_1

and

P_3

have the same complex roots since they have the same vertex x-coordinate

P_3

has one real root and no complex roots, while

P_1

has two complex roots

P_1

has two distinct complex roots while

P_3

has one repeated real root

The complex roots of

P_1

have non-zero imaginary parts while

P_3

has a double root with zero imaginary part

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Math 2 Quiz

Math 2 Quiz: No Real Intercepts And Complex Roots

Practice No Real Intercepts And Complex Roots in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on No Real Intercepts And Complex Roots, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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All questions

Question 1

$P_1$ and $P_3$ have the same complex roots since they have the same vertex x-coordinate
$P_3$ has one real root and no complex roots, while $P_1$ has two complex roots
$P_1$ has two distinct complex roots while $P_3$ has one repeated real root
The complex roots of $P_1$ have non-zero imaginary parts while $P_3$ has a double root with zero imaginary part (correct answer)

Explanation: 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.

Question 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?

For all real $t$ , there are no real x-intercepts, and the complex roots are always $-t \pm 2i$ (correct answer)
For $|t| > 2$ , there are no real x-intercepts, and the complex roots have real parts equal to $-t$
For $|t| < 2$ , there are no real x-intercepts, and the imaginary parts of the roots depend on $t$
For all real $t$ , there are real x-intercepts except when $t = \pm 2$ , where there are repeated real roots

Explanation: 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.

Question 3

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?

The value $d$ can be any real number since parabolas always intersect horizontal lines at complex points when extended to the complex plane
The value $d$ equals the y-coordinate of the vertex, making the intersection a single point of tangency
The value $d$ is greater than the minimum value of the parabola when $a > 0$ , or less than the maximum value when $a < 0$
The value $d$ is less than the minimum value of the parabola when $a > 0$ , or greater than the maximum value when $a < 0$ (correct answer)

Explanation: 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.

Question 4

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?

Yes, because the y-intercept equals the product of the distances from each root to the origin
Yes, because $c = a(p^2 + q^2)$ and when $a = 1$ , the y-intercept equals $p^2 + q^2$
No, because the y-intercept is $c = a(p^2 + q^2)$ , which equals $p^2 + q^2$ only when $a = 1$ (correct answer)
No, because there is no general relationship between the y-intercept and the modulus of complex roots

Explanation: 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.

Question 5

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?

$3 \pm 2i$ (correct answer)
$-3 \pm 2i$
$5 \pm 3i$
$3 \pm 5i$

Explanation: 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.

Question 6

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 $q(x)$ ?

$q(x)$ must have two distinct real zeros since the discriminant changes when the leading coefficient changes sign (correct answer)
$q(x)$ must have no real zeros since it shares the same linear and constant terms as $p(x)$
$q(x)$ must have exactly one real zero since the parabolas are reflections of each other
$q(x)$ must have no real zeros since both parabolas avoid the x-axis completely

Explanation: 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.

Question 7

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?

$k > -4$
$k > 0$
$k > 4$ (correct answer)
$k > 8$

Explanation: 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.

Question 8

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?

$0$ (correct answer)
$\pm 2i$
$4i$
$\pm 4i$

Explanation: 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.

Question 9

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?

The graph intersects the y-axis below the x-axis and has a minimum point at $(3, 0)$
The graph intersects the y-axis above the x-axis and has a minimum point above the x-axis (correct answer)
The graph intersects the y-axis above the x-axis and has a maximum point below the x-axis
The graph intersects the y-axis below the x-axis and has a maximum point at $(3, -26)$

Explanation: 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.

Question 10

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?

Both functions have complex roots with the same real parts since their vertices have the same x-coordinate
The function $f(x)$ has complex roots with larger imaginary parts because its discriminant is more negative
The function $g(x)$ has complex roots with larger imaginary parts because its discriminant is more negative (correct answer)
Both functions have the same imaginary parts in their complex roots since both discriminants equal $-12$

Explanation: 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).

Question 11

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?

They have identical complex roots since one function is just a vertical stretch of the other
They have the same complex roots because vertical stretching doesn't affect x-intercepts (correct answer)
They have different complex roots, but the real parts are the same while imaginary parts differ by a factor of $\sqrt{2}$
They have different complex roots, with $f(x)$ having roots that are exactly twice those of $g(x)$

Explanation: 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.

Question 12

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 $f(x) = 0$ ?

The roots of $g(x) = 0$ are $r_1 + 3$ and $r_2 + 3$
The roots of $g(x) = 0$ are $r_1 - 3$ and $r_2 - 3$ (correct answer)
The roots of $g(x) = 0$ are the same as those of $f(x) = 0$ because horizontal shifts don't affect roots
The roots of $g(x) = 0$ have the same imaginary parts as $f(x) = 0$ but different real parts

Explanation: 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.

Question 13

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?

The graph has a maximum point below the x-axis, confirming that both complex roots have negative real parts
The graph has a maximum point above the x-axis, but the function still has complex roots with positive real parts
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 (correct answer)
The graph has a minimum point above the x-axis, so the complex roots must have imaginary parts of equal magnitude

Explanation: 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.