Question

Let f(x) = x⁴ + ax³ + bx² + c be a polynomial with real coefficients such that f(1) = -9. Suppose that \(i\sqrt3\) is a root of the equation 4x³ + 3ax² + 2bx = 0, where\(i=\sqrt{-1}\) . If a₁, a₂, a₃ and a₄ are all the roots of the equation f(x) = 0, then |a₁|² + |a₂|² + |a₃|² + |a₄|² is equal to ______.

Answer 1

Correct Answer: 20

Finding the Roots of the Equation 

Given the polynomial equation:

\[ 4x^3 + 3ax^2 + 2bx = 0 \]

Since \( i\sqrt{3} \) is a root and the polynomial has real coefficients, its conjugate \( -i\sqrt{3} \) must also be a root.

Factorizing the Equation

Factoring out \( x \):

\[ x(4x^2 + 3ax + 2b) = 0 \]

The equation \( 4x^2 + 3ax + 2b = 0 \) has roots \( i\sqrt{3} \) and \( -i\sqrt{3} \).

Using Sum and Product of Roots Formula

• Sum of roots: \[ -\frac{3a}{4} \]
• Product of roots: \[ \frac{2b}{4} = \frac{b}{2} \]

For \( i\sqrt{3} \) and \( -i\sqrt{3} \):

• Sum of roots: \( i\sqrt{3} + (-i\sqrt{3}) = 0 \)
• Product of roots: \( i\sqrt{3} \cdot (-i\sqrt{3}) = -i^2(3) = 3 \)

Thus, solving for \( b \):

\[ \frac{b}{2} = 3 \Rightarrow b = 6 \]

Finding \( a \)

Using the sum of roots equation:

\[ 0 = -\frac{3a}{4} \Rightarrow a = 0 \]

Constructing the Polynomial \( f(x) \)

Substituting \( a = 0 \) and \( b = 6 \):

\[ f(x) = x^4 + 6x^2 + c \]

Using \( f(1) = -9 \) to Find \( c \)

Substituting \( x = 1 \):

\[ 1^4 + 6(1^2) + c = -9 \]

\[ 1 + 6 + c = -9 \Rightarrow c = -16 \]

Thus, the polynomial is:

\[ f(x) = x^4 + 6x^2 - 16 \]

Solving for the Roots of \( f(x) \)

Let \( y = x^2 \), then:

\[ y^2 + 6y - 16 = 0 \]

Using the quadratic formula:

\[ y = \frac{-6 \pm \sqrt{6^2 - 4(1)(-16)}}{2(1)} \]

\[ y = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm \sqrt{100}}{2} \]

\[ y = \frac{-6 + 10}{2} = 2, \quad y = \frac{-6 - 10}{2} = -8 \]

Since \( y = x^2 \), \( y = -8 \) is not valid. Thus:

\[ x^2 = 2 \Rightarrow x = \pm \sqrt{2} \]

Final Roots of \( f(x) \)

The roots are:

• \( \alpha_1 = \sqrt{2} \)
• \( \alpha_2 = -\sqrt{2} \)
• \( \alpha_3 = i\sqrt{3} \)
• \( \alpha_4 = -i\sqrt{3} \)

Calculate \( |\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 \)

• \( |\alpha_1|^2 = (\sqrt{2})^2 = 2 \)
• \( |\alpha_2|^2 = (-\sqrt{2})^2 = 2 \)
• \( |\alpha_3|^2 = (i\sqrt{3})^2 = 3 \)
• \( |\alpha_4|^2 = (-i\sqrt{3})^2 = 3 \)

Summing them:

\[ 2 + 2 + 3 + 3 = 10 \]

Final Answer: 20