For a real number a, the quadratic equation x^2 + ax + (a - 3)^2 = 0 has complex roots. On… — ACT

For a real number $a$ , the quadratic equation $x^2 + ax + (a - 3)^2 = 0$ has complex roots. One of these roots is $2 + i\sqrt{b}$ , where $b$ is a positive real number and $i = \sqrt{-1}$ . What is the value of $a + b$ ?

A
17
B
-21
41Answer
D
-49
E
1

Answer

The value of

a + b

41

The correct answer is

41

. Since the quadratic equation has real coefficients, its complex roots must occur in conjugate pairs. Thus, the roots are

2 + i\sqrt{b}

and

2 - i\sqrt{b}

. Their sum is

4

, which by Vieta's formulas equals

-a

, giving

a = -4

. The product of the roots is

4 + b

, which equals the constant term

(a - 3)^2 = (-4 - 3)^2 = 49

. Solving for

b

gives

b = 45

. Therefore,

a + b = -4 + 45 = 41

Step-by-Step Solution

Identify the second root using the complex conjugate root theorem.

The conjugate root is

2 - i\sqrt{b}

Since the coefficients of the quadratic equation are real, complex roots must occur in conjugate pairs.

Find the value of

a

using the sum of the roots.

a = -4

By Vieta's formulas, the sum of the roots is equal to

-a

. The sum of the conjugate roots is

(2 + i\sqrt{b}) + (2 - i\sqrt{b}) = 4

, so

-a = 4

Calculate the constant term of the quadratic equation.

The constant term is

49

Substitute

a = -4

into the constant term

(a - 3)^2

to get

(-4 - 3)^2 = (-7)^2 = 49

Find the value of

b

using the product of the roots.

b = 45

By Vieta's formulas, the product of the roots is the constant term. The product is

(2 + i\sqrt{b})(2 - i\sqrt{b}) = 4 - i^2 b = 4 + b

(since

i^2 = -1

). Equating this to the constant term gives

4 + b = 49

, so

b = 45

Calculate the final value of

a + b

41

Substitute the values of

a

and

b

a + b = -4 + 45 = 41

Key Concept

Vieta's formulas and complex conjugate roots of quadratic equations with real coefficients.

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