Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20208 Jan 2020Morning ShiftMathematicsComplex NumberActual

If the equation x 2 + b x + 45 = 0 , b ∈ R has conjugate complex roots and they satisfy z + 1 = 2 10 , then

Options

  1. Ab 2 - b = 30
  2. Bb 2 + b = 72
  3. Cb 2 - b = 42
  4. Db 2 + b = 12

Correct answer

A. b 2 - b = 30

Step-by-step solution

Let z = α + i β be one of the roots of the equation x 2 + b x + 45 = 0 ,   b ∈ R . So, its other conjugate complex root will be z = α + i β ¯ = α - i β . We know that for a quadratic equation A x 2 + B x + C = 0 , the sum and product of its roots are - B A   &   C A respectively. So, the sum of roots of the given equation, α + i β + α - i β = - b 1 ⇒ 2 α = - b   . . . . . . i . Also, the product of roots of the given equ

Practice Complex Number on Quantrex Academy →

More from Complex Number

The number of values of z C , satisfying the equations |z-(4+8i)|= 10 and |z-(3+5i)|+|z-(5+11i)|=4 5 , is: 2026Let S = z C : z^2 + 6 ,iz - 3 = 0 . Then _ z S z^8 is equal to : 2026Let the set of all values of k R such that the equation z( z + 2 + i) + k(2 + 3i) = 0 , z C , has at least one solution, be the interval [ , ] . Then 9( + ) is equal to: 2026Let z₁, z₂ C be the distinct solutions of the equation z^2 + 4z - (1 + 12i) = 0 . Then |z₁|^2 + |z₂|^2 is equal to : 2026Let S= z C : z^2+4z+16=0 . Then _ z S |z+ 3 i|^2 is equal to: 2026Let z be a complex number such that |z+2| = |z-2| and ( z+3 z-i ) = 4 . Then |z|^2 is equal to: 2026Let the circles C₁ : |z| = r and C₂ : |z - 3 - 4i| = 5 , z C , be such that C₂ lies within C₁ . If z₁ moves on C₁ , z₂ moves on C₂ and |z₁ - z₂| = 2 , then |z₁ - z₂| is equal to: 2026Let x and y be real numbers such that 50 ( 2x 1+3i - y 1-2i ) = 31 + 17i , i = -1 . Then the value of 10(x - 3y) is : 2026 Full Complex Number list All JEE Main PYQs