JEE Advanced 2024 — Paper 1 — MATH-III solution

JEE Advanced2024Paper 1MATH-III

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

Your answer

Solution

From $f (1) = - 9$ : $1 + a + b + c = - 9$ , i.e. $a + b + c = - 10$ .

The derivative is $f^{'} (x) = 4 x^{3} + 3 a x^{2} + 2 b x = x (4 x^{2} + 3 a x + 2 b)$ . Given $i 3$ is a root, by real-coefficient conjugate pairing $- i 3$ is also a root of $4 x^{2} + 3 a x + 2 b = 0$ .

Sum of roots: $i 3 + (- i 3) = 0 = - \frac{3 a}{4} ⟹ a = 0$ .

Product of roots: $(i 3) (- i 3) = 3 = \frac{2 b}{4} ⟹ b = 6$ .

Then $c = - 10 - 6 = - 16$ , so $f (x) = x^{4} + 6 x^{2} - 16 = 0$ .

Setting $u = x^{2}$ : $u^{2} + 6 u - 16 = 0 ⟹ u = \frac{- 6 \pm 36 + 64}{2} = 2, - 8$ .

The four roots are $\pm 2, \pm 22 i$ . Sum of $∣ α_{k} ∣^{2}$ :

$2 + 2 + 8 + 8 = 20$

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