For what value of the constant c does the quadratic equation x^2 - 3x + c = 0 have two com… — ACT

For what value of the constant $c$ does the quadratic equation $x^2 - 3x + c = 0$ have two complex solutions with imaginary parts equal to $\pm 2i$ ?

The value of the constant

c

6.25

The correct answer is

6.25

. Applying the quadratic formula to

x^2 - 3x + c = 0

gives solutions of the form

1.5 \pm \frac{\sqrt{9 - 4c}}{2}

. Since these solutions are complex with imaginary parts equal to

\pm 2i

, the term under the radical must be negative, and the imaginary component is

\frac{\sqrt{9 - 4c}}{2} = 2i

. Multiplying both sides by

2

gives

\sqrt{9 - 4c} = 4i

. Squaring both sides results in

9 - 4c = 16i^2

. Substituting

i^2 = -1

gives

9 - 4c = -16

. Solving for

c

yields

-4c = -25

, which simplifies to

c = 6.25

Apply the quadratic formula to the equation

x^2 - 3x + c = 0

The solutions are given by

x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(c)}}{2(1)} = 1.5 \pm \frac{\sqrt{9 - 4c}}{2}

This expresses the solutions in terms of the constant

c

so that the imaginary part can be identified.

Set the imaginary term of the solutions equal to the given imaginary parts

\pm 2i

\frac{\sqrt{9 - 4c}}{2} = 2i \implies \sqrt{9 - 4c} = 4i

The question specifies that the imaginary parts of the two complex solutions are

\pm 2i

Square both sides of the equation to solve for

c

9 - 4c = (4i)^2 = 16i^2

. Since

i^2 = -1

, this becomes

9 - 4c = -16

Squaring eliminates the radical and allows for standard algebraic isolation of the variable

c

Solve the linear equation for

c

-4c = -25 \implies c = 6.25

Subtracting

9

from both sides and then dividing by

-4

isolates the constant

c

Solving quadratic equations with complex roots using the quadratic formula and the properties of the imaginary unit.

Question