Question

Let $f(x)=x^4+ax^3+bx^2+cx+d$ be a polynomial with real coefficients such that $f(1)=-9$. Suppose that $i\sqrt{3}$ is a root of the equation $x^3+2x^2+4x+3=0$, where $i^2=-1$. If $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ are all the roots of the equation $f(x)=0$, then $\alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2$ is equal to ______.

Answer 1

161/100

Answer 2

Step 1: Find the roots of the cubic
The given cubic is

$$x^3 + 2x^2 + 4x + 3 = 0.$$

Since $i\sqrt{3}$ is a root and coefficients are real, the conjugate $-\,i\sqrt{3}$ is also a root. Let the third real root be $r$.
By Viète’s formula, the sum of roots is

$$(i\sqrt{3}) + (-\,i\sqrt{3}) + r = -2 \quad\Longrightarrow\quad r = -2.$$

So the roots are $i\sqrt{3},\,-i\sqrt{3},\,-2$.

Step 2: Construct $f(x)$
Since these three are roots of $f(x)$, write

$$f(x) = (x^3 + 2x^2 + 4x + 3)\,(x - k),$$

for some real $k$. Expanding:

$$f(x) = x^4 + (2 - k)x^3 + (4 - 2k)x^2 + (3 - 4k)x - 3k.$$

Now use $f(1)=-9$:

$$f(1)=1+(2-k)+(4-2k)+(3-4k)-3k = 10 - 10k = -9 \quad\Longrightarrow\quad 10 - 10k = -9 \quad\Longrightarrow\quad k = \tfrac{19}{10}.$$

Step 3: Identify all four roots
Thus the four roots of $f(x)$ are

$$\alpha_1 = i\sqrt{3},\quad \alpha_2 = -\,i\sqrt{3},\quad \alpha_3 = -2,\quad \alpha_4 = \tfrac{19}{10}.$$

Step 4: Compute the desired sum

$$\sum_{j=1}^4 \alpha_j^2 = (i\sqrt{3})^2 + (-\,i\sqrt{3})^2 + (-2)^2 + \bigl(\tfrac{19}{10}\bigr)^2 = (-3) + (-3) + 4 + \tfrac{361}{100} = -2 + \tfrac{361}{100} = \tfrac{161}{100}.$$