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

Let \(f(x) = x^4 + ax^3 + bx^2 + c\) be a polynomial with real coefficients. 
Given that \(f(1) = -9\), so \(1^4 + a(1)^3 + b(1)^2 + c = -9\), which gives \(1 + a + b + c = -9\), or \(a + b + c = -10\). 
Also, \(i\sqrt{3}\) is a root of \(4x^3 + 3ax^2 + 2bx = 0\). 
Since \(i\sqrt{3}\) is a root, \(-i\sqrt{3}\) is also a root. 
So \(4(i\sqrt{3})^3 + 3a(i\sqrt{3})^2 + 2b(i\sqrt{3}) = 0\). \(4(i^3 3\sqrt{3}) + 3a(-3) + 2bi\sqrt{3} = 0\). \(4(-i3\sqrt{3}) - 9a + 2bi\sqrt{3} = 0\). \(-12i\sqrt{3} - 9a + 2bi\sqrt{3} = 0\). 
Equating the real and imaginary parts to 0, we get: \(-9a = 0 \Rightarrow a = 0\). 
\(-12\sqrt{3} + 2b\sqrt{3} = 0 \Rightarrow 2b = 12 \Rightarrow b = 6\). Now, \(a + b + c = -10\), so \(0 + 6 + c = -10\), which gives \(c = -16\). Then \(f(x) = x^4 + 6x^2 - 16\). \(f(x) = (x^2 - 2)(x^2 + 8) = 0\). 
The roots of \(f(x)\) are \(x = \pm \sqrt{2}\) and \(x = \pm i\sqrt{8} = \pm 2i\sqrt{2}\). 
Let \(a_1 = \sqrt{2}, a_2 = -\sqrt{2}, a_3 = 2i\sqrt{2}, a_4 = -2i\sqrt{2}\). 
\(|a_1|^2 = (\sqrt{2})^2 = 2\). \(|a_2|^2 = (-\sqrt{2})^2 = 2\). 
\(|a_3|^2 = (2i\sqrt{2})^2 = (0+2\sqrt{2}i)(0-2\sqrt{2}i) = (2\sqrt{2})^2 = 8\). 
\(|a_4|^2 = (-2i\sqrt{2})^2 = (0-2\sqrt{2}i)(0+2\sqrt{2}i) = (2\sqrt{2})^2 = 8\). 
Then \(|a_1|^2 + |a_2|^2 + |a_3|^2 + |a_4|^2 = 2 + 2 + 8 + 8 = 20\).

Final Answer: The final answer is \(\boxed{20}\)

Answer 2

To solve the problem, analyze the polynomial \(f(x) = x^4 + a x^3 + b x^2 + c\) and its roots, using given conditions.

Given:
- \(f(1) = -9\)
- \(i \sqrt{3}\) is a root of \(4x^3 + 3 a x^2 + 2 b x = 0\)
- Roots of \(f(x) = 0\) are \(a_1, a_2, a_3, a_4\)
- Need to find \(|a_1|^2 + |a_2|^2 + |a_3|^2 + |a_4|^2\)

Step 1: Understand the relation between \(f(x)\) and the given cubic:
Note that \[ f'(x) = \frac{d}{dx} f(x) = 4 x^3 + 3 a x^2 + 2 b x \] Given that \(i \sqrt{3}\) is a root of \(f'(x) = 0\).

Step 2: Since coefficients of \(f\) are real, roots of \(f'(x)\) come in conjugate pairs. So, if \(i \sqrt{3}\) is a root, then \(-i \sqrt{3}\) is also a root of \(f'(x) = 0\).

Step 3: Express \(f'(x)\) using known roots:
Let the roots of \(f'(x)\) be \(r_1, r_2, r_3\). Two roots are \(i \sqrt{3}\) and \(-i \sqrt{3}\), so:
\[ f'(x) = 4(x - i \sqrt{3})(x + i \sqrt{3})(x - r_3) = 4 (x^2 + 3)(x - r_3) \] Expanding: \[ 4 (x^3 - r_3 x^2 + 3 x - 3 r_3) \] Comparing with given expression: \[ 4 x^3 + 3 a x^2 + 2 b x \] Match coefficients: \[ 4 x^3 - 4 r_3 x^2 + 12 x - 12 r_3 = 4 x^3 + 3 a x^2 + 2 b x + 0 \] This gives: \[ -4 r_3 = 3 a \implies r_3 = -\frac{3 a}{4} \] \[ 12 = 2 b \implies b = 6 \] \[ -12 r_3 = 0 \implies r_3 = 0 \] But we got two different values for \(r_3\), so the only consistent solution is: \[ r_3 = 0 \] Thus, \[ -4 r_3 = 3 a \implies 0 = 3 a \implies a = 0 \] And, \[ b = 6 \]

Step 4: Now, \(f(x) = x^4 + 0 \cdot x^3 + 6 x^2 + c = x^4 + 6 x^2 + c\)

Step 5: Use \(f(1) = -9\):
\[ 1^4 + 6 \times 1^2 + c = -9 \implies 1 + 6 + c = -9 \implies c = -16 \] So, \[ f(x) = x^4 + 6 x^2 - 16 \]

Step 6: Factorize \(f(x) = x^4 + 6 x^2 -16\):
Try factorization as: \[ (x^2 + p)(x^2 + q) = x^4 + (p + q) x^2 + p q \] Comparing, \[ p + q = 6, \quad p q = -16 \] Solve quadratic for \(p\): \[ p^2 - 6 p -16 = 0 \] \[ p = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm 10}{2} \] \[ p = 8 \quad \text{or} \quad p = -2 \] So factors are: \[ (x^2 + 8)(x^2 - 2) \] Roots are: \[ a_{1,2} = \pm i \sqrt{8} = \pm i 2 \sqrt{2}, \quad a_{3,4} = \pm \sqrt{2} \]

Step 7: Calculate sum of squared magnitudes:
\[ |a_1|^2 + |a_2|^2 = (2 \sqrt{2})^2 + (2 \sqrt{2})^2 = 8 + 8 = 16 \] \[ |a_3|^2 + |a_4|^2 = (\sqrt{2})^2 + (-\sqrt{2})^2 = 2 + 2 = 4 \] \[ \text{Total} = 16 + 4 = 20 \]

Final Answer:
\[ \boxed{20} \]