Let f(x) = x⁴ + ax³ + bx² + c be a polynomial with real coefficients such that f(1) = -9. Suppose th

Q: Wait, why is the auxiliary equation $4x^{3} + 3ax^{2} + 2bx = 0$ actually $f'(x) = 0$?

Great observation! If you differentiate $f(x) = x^{4} + ax^{3} + bx^{2} + c$ with respect to $x$, you get $f'(x) = 4x^{3} + 3ax^{2} + 2bx$. The problem gives you this derivative set to zero. Recognizing this connection saves you from blindly guessing and lets you use the properties of calculus to solve for the coefficients.

Q: Why can't we just use Vieta's formulas immediately?

Vieta's formulas are powerful, but they relate the roots to the coefficients. Here, we don't know the coefficients $a, b,$ and $c$ yet! We must first find these coefficients using the given root and the value of $f(1)$. Once the polynomial is fully defined, you could use Vieta's, but solving for the roots directly is often more intuitive for biquadratic equations.

Q: Why do we reject the real root for the imaginary part?

We don't reject it; we categorize it! When we solve $x^{2} = -8$, we get $x = \pm \sqrt{-8}$. Since the square of any real number is non-negative, there are no real solutions for $x^{2} = -8$. This forces us into the complex plane, where $x = \pm i\sqrt{8} = \pm 2i\sqrt{2}$. It's not a rejection; it's an expansion of our mathematical horizon.

Question

Let $f(x) = x^{4} + ax^{3} + bx^{2} + c$ be a polynomial with real coefficients such that $f(1) = -9$. Suppose that $i\sqrt{3}$ is a root of the equation $4x^{3} + 3ax^{2} + 2bx = 0$, where $i = \sqrt{-1}$. If $a_{1}, a_{2}, a_{3}$, and $a_{4}$ are all the roots of the equation $f(x) = 0$, then $|a_{1}|^{2} + |a_{2}|^{2} + |a_{3}|^{2} + |a_{4}|^{2}$ is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dissect a problem that might look like a standard algebra exercise, but beneath the surface, it is a beautiful dance of complex numbers and calculus.

We are given a polynomial $f(x) = x^{4} + ax^{3} + bx^{2} + c$ and an auxiliary equation $4x^{3} + 3ax^{2} + 2bx = 0$.

If you take the derivative of $f(x)$, you get $f'(x) = 4x^{3} + 3ax^{2} + 2bx$. The auxiliary equation is simply $f'(x) = 0$. This is the first layer of the mystery peeled back.

### The Complex Dance

We are told that $i\sqrt{3}$ is a root of this auxiliary equation. Since the coefficients $a$ and $b$ are real, we know that complex roots must appear in conjugate pairs. This means $-i\sqrt{3}$ is also a root.

When we plug $x = i\sqrt{3}$ into $4x^{3} + 3ax^{2} + 2bx = 0$, we must be meticulous. Remember that $i^{2} = -1$ and $i^{3} = -i$. Thus, $(i\sqrt{3})^{2} = -3$ and $(i\sqrt{3})^{3} = -3i\sqrt{3}$.

Substituting these into our equation yields:

$$4(-3i\sqrt{3}) + 3a(-3) + 2b(i\sqrt{3}) = 0$$

This simplifies to $-12i\sqrt{3} - 9a + 2bi\sqrt{3} = 0$. Grouping the real and imaginary parts, we get:

$$-9a + i\sqrt{3}(2b - 12) = 0$$

For this to hold true, both parts must vanish independently. This gives us the elegant results **$a = 0$** and **$b = 6$**.

### The Biquadratic Reveal

With $a$ and $b$ in hand, our polynomial transforms. It is no longer a mystery; it is $f(x) = x^{4} + 6x^{2} + c$.

We are given $f(1) = -9$. Substituting $x=1$, we get $1 + 6 + c = -9$, which leads us directly to **$c = -16$**. Our polynomial is now fully revealed:

$$f(x) = x^{4} + 6x^{2} - 16$$

To find the roots $a_{1}, a_{2}, a_{3}, a_{4}$, we set $f(x) = 0$. By substituting $y = x^{2}$, we reduce it to the quadratic $y^{2} + 6y - 16 = 0$.

Factoring this is straightforward: $(y + 8)(y - 2) = 0$. This gives us $y = 2$ and $y = -8$.

### The Final Calculation

Now, we return to $x$. If $x^{2} = 2$, then $x = \pm \sqrt{2}$. These are our real roots, $a_{1}$ and $a_{2}$.

If $x^{2} = -8$, then $x = \pm \sqrt{-8} = \pm 2i\sqrt{2}$. These are our imaginary roots, $a_{3}$ and $a_{4}$.

The problem asks for the sum of the squares of the moduli: $|a_{1}|^{2} + |a_{2}|^{2} + |a_{3}|^{2} + |a_{4}|^{2}$.

The modulus of $\pm \sqrt{2}$ is $\sqrt{2}$, so its square is $2$. The modulus of $\pm 2i\sqrt{2}$ is $2\sqrt{2}$, so its square is $8$.

Summing these up, we get:

$$2 + 2 + 8 + 8 = 20$$

You have successfully navigated the complex plane and the algebraic landscape. The final answer is **$20$**.

Enter Numerical Value:

Visualized Solution

Given Polynomial and Equation

Substituting the Root

Expanding the Powers

Grouping Real and Imaginary Parts

Solving for $a$ and $b$

Updating the Polynomial

Solving for $c$

Finding the Roots of $f (x)$

Solving the Quadratic Equation

Finding the Real Roots

Finding the Imaginary Roots

Calculating the Moduli Squares

Final Summation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Complex Dance

The Biquadratic Reveal

The Final Calculation

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