Complex Solutions of Quadratics Worksheet Answer Key – Algebra 2

Problem 1 Discriminant and type of solutions

a) D = (−3)² − 4(1)(−10) = 49: two distinct real solutions.

b) D = (−8)² − 4(1)(16) = 0: one repeated real solution.

c) D = (3)² − 4(2)(4) = −23: two non-real complex solutions (a conjugate pair).

d) D = (−2)² − 4(3)(−7) = 88: two distinct real solutions.

Problem 2 Solving with square roots

a) x = ±√(−49) = ±7i.

b) x² = −36, so x = ±6i.

c) 3x² = −75, x² = −25, so x = ±5i.

Problem 3 Solutions from a squared binomial

a) x + 4 = ±5i, so x = −4 ± 5i. −4 + 5i and −4 − 5i

b) (x − 3)² = −49, so x − 3 = ±7i and x = 3 ± 7i.

Problem 4 Quadratic formula with a negative discriminant

Here a = 1, b = 10, c = 41.

Discriminant: D = b² − 4ac = (10)² − 4(1)(41) = −64.

D < 0, so there are two non-real solutions. √(−64) = 8i.

x = −(10) ± √(−64)2(1)

x = −5 ± 4i

Problem 5 Reading the graph

The vertex (4, 3) is above the x-axis and the parabola opens upward, so the graph never crosses the axis: 0 real solutions.

Here a = 1, b = −8, c = 19.

Discriminant: D = b² − 4ac = (−8)² − 4(1)(19) = −12.

D < 0, so there are two non-real solutions. √(−12) = 2i√3.

x = −(−8) ± √(−12)2(1)

x = 4 ± i√3

Problem 6 Completing the square

x² − 12x = −45. Add 36 to both sides: (x − 6)² = −9.

x − 6 = ±3i, so x = 6 ± 3i.

Problem 7 Fractions in the solution

Here a = 3, b = 5, c = 4.

Discriminant: D = b² − 4ac = (5)² − 4(3)(4) = −23.

D < 0, so there are two non-real solutions. √(−23) = i√23.

x = −(5) ± √(−23)2(3)

x = −5 ± i√236

Problem 8 A parabola that opens downward

The vertex (1, −5) is below the axis and the parabola opens downward, so it never reaches y = 0.

Multiply by −1: x² − 2x + 6 = 0.

Here a = 1, b = −2, c = 6.

Discriminant: D = b² − 4ac = (−2)² − 4(1)(6) = −20.

D < 0, so there are two non-real solutions. √(−20) = 2i√5.

x = −(−2) ± √(−20)2(1)

x = 1 ± i√5

Problem 9 A solution with a reduced fraction

Here a = 5, b = −4, c = 1.

Discriminant: D = b² − 4ac = (−4)² − 4(5)(1) = −4.

D < 0, so there are two non-real solutions. √(−4) = 2i.

x = −(−4) ± √(−4)2(5)

x = 2 ± i5

Problem 10 Reducing a radical solution

Here a = 3, b = −6, c = 10.

Discriminant: D = b² − 4ac = (−6)² − 4(3)(10) = −84.

D < 0, so there are two non-real solutions. √(−84) = 2i√21.

x = −(−6) ± √(−84)2(3)

x = 3 ± i√213

Problem 11 Choosing a constant

D = (−8)² − 4(1)k = 64 − 4k.

Non-real solutions need D < 0: 64 − 4k < 0, so k > 16.

One repeated real solution needs D = 0: k = 16.

k > 16; k = 16

Problem 12 Plotting complex solutions

Here a = 1, b = 6, c = 18.

Discriminant: D = b² − 4ac = (6)² − 4(1)(18) = −36.

D < 0, so there are two non-real solutions. √(−36) = 6i.

x = −(6) ± √(−36)2(1)

x = −3 ± 3i

Sum: (−3 + 3i) + (−3 − 3i) = −6 = −(6)/1. Product: 9 − 9i² = 18 = 18/1.

−3 + 3i and −3 − 3i; sum −6, product 18

Problem 13 Building an equation from solutions

Sum = 14 and product = 49 + 4 = 53.

x² − (sum)x + (product) = 0.

x² − 14x + 53 = 0

Problem 14 Which equations have non-real solutions

a) D = −3: non-real.

b) D = 49: real.

c) D = 0: real.

d) D = −47: non-real.

Non-real solutions: (a) and (d).

Problem 15 Completing the square with a leading coefficient

a) x² + 6x = −25, so (x + 3)² = −16 and x + 3 = ±4i. x = −3 ± 4i

b) Divide by 2: x² − 6x + 13 = 0. Then (x − 3)² = −4, x − 3 = ±2i. x = 3 ± 2i

Problem 16 Will the ball reach the height?

Set −16t² + 40t + 5 = 40, so 16t² − 40t + 35 = 0.

Here a = 16, b = −40, c = 35.

Discriminant: D = b² − 4ac = (−40)² − 4(16)(35) = −640.

D < 0, so there are two non-real solutions. √(−640) = 8i√10.

t = −(−40) ± √(−640)2(16)

t = 5 ± i√104

D < 0, so the ball never reaches 40 feet; its maximum is 5 + 25 = 30 feet.

Problem 17 Profit that cannot happen

Set −3n² + 36n − 90 = 100, so 3n² − 36n + 190 = 0.

D = (−36)² − 4(3)(190) = 1296 − 2280 = −984 < 0, so there is no real n.

The vertex is at n = 6: P(6) = −108 + 216 − 90 = 18.

Profit never reaches $100; the maximum is $18

Problem 18 Sum and product of solutions

Here a = 2, b = −10, c = 17.

Discriminant: D = b² − 4ac = (−10)² − 4(2)(17) = −36.

D < 0, so there are two non-real solutions. √(−36) = 6i.

x = −(−10) ± √(−36)2(2)

x = 5 ± 3i2

Sum: 5 + 3i2 + 5 − 3i2 = 5, and −b/a = 10/2 = 5.

Product: 25 − 9i²4 = 344 = 172 = c/a.

Sum 5, product 172; both match

Problem 19 Ranges of a parameter

D = k² − 64.

a) D < 0 when k² < 64, that is −8 < k < 8.

b) D = 0 when k = 8 or k = −8.

c) D > 0 when k < −8 or k > 8.

(a) −8 < k < 8; (b) k = ±8; (c) |k| > 8

Problem 20 Finding a missing coefficient

Real coefficients force the conjugate 6 + 4i to be the other solution.

Sum = 12, so −b = 12 and b = −12. Product = 36 + 16 = 52, so c = 52.

Other solution 6 + 4i; b = −12, c = 52