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← JEE Advanced 2024 · Paper 1
MathematicsInteger+4Question 9 of 51

JEE Advanced 2024 · Paper 1, Mathematics question 9

Let f(x)=x4+ax3+bx2+cf(x) = x^4 + ax^3 + bx^2 + c be a polynomial with real coefficients such that f(1)=−9f(1) = -9. Suppose that i3i\sqrt{3} is a root of the equation 4x3+3ax2+2bx=04x^3 + 3ax^2 + 2bx = 0, where i=−1i = \sqrt{-1}. If α1,α2,α3\alpha_1, \alpha_2, \alpha_3, and α4\alpha_4 are all the roots of the equation f(x)=0f(x) = 0, then ∣α1∣2+∣α2∣2+∣α3∣2+∣α4∣2|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 is equal to _____.

Answer and worked solution

Official answer: 20

f′(x)=4x3+3ax2+2bx=x(4x2+3ax+2b)f'(x) = 4x^3 + 3ax^2 + 2bx = x(4x^2 + 3ax + 2b). The coefficients are real, so the roots ±i3\pm i\sqrt3 are both roots of the quadratic factor:

  • sum of roots 0=−3a4  ⟹  a=00 = -\frac{3a}{4} \implies a = 0;
  • product (i3)(−i3)=3=2b4  ⟹  b=6(i\sqrt3)(-i\sqrt3) = 3 = \frac{2b}{4} \implies b = 6.

f(1)=1+6+c=−9  ⟹  c=−16f(1) = 1 + 6 + c = -9 \implies c = -16. So
f(x)=x4+6x2−16=(x2+8)(x2−2),f(x) = x^4 + 6x^2 - 16 = (x^2+8)(x^2-2),
with roots ±22 i\pm 2\sqrt2\,i and ±2\pm\sqrt2.

∑∣αk∣2=8+8+2+2=20\sum|\alpha_k|^2 = 8 + 8 + 2 + 2 = 20.

Topics:Theory of equations (higher degree)Modulus and argument
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