JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

If the equation, x² + bx + 45 = 0 (b ∈ R) has conjugate complex roots and they satisfy |z +1| = 2√10 , then :

  1. (a) b² – b = 42
  2. (b) b² + b = 12
  3. (c) b² + b = 72
  4. (d) b² – b = 30
Answer and solution

Answer (d) b² – b = 30

Solution

Maths sir · 4 steps

Do not solve for the roots. Expand |z+1|², and the sum and product of the roots finish the job.

  1. Use Vieta. The roots are z and z̄. Their product is zz̄ = |z|² = 45 and their sum is z + z̄ = -b.

    |z+1|² = (z+1)(z̄+1) = zz̄ + (z+z̄) + 1

  2. Put in values. The right side is 45 - b + 1 = 46 - b. The left side is (2√10)² = 40. So 46 - b = 40, which gives b = 6.
  3. Check the roots. The discriminant is b² - 180 = 36 - 180 < 0, so the roots really are complex conjugates and b = 6 is valid.
  4. So. b² - b = 36 - 6 = 30, which is option (d).

The idea to keep. When a quadratic with real coefficients has roots z, z̄, write |z+a|² = (z+a)(z̄+a). Then the sum and product of the roots give the answer without finding z.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Complex Numbers

Asked about 160 times in JEE Main since 2020. This question is from 2020.