Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main2015MathematicsComplex NumberActual

If 2 + 3 i is one of the roots of the equation 2 x 3 - 9 x 2 + k x - 13 = 0 , k ∈ R , then the real root of this equation (where i 2 = - 1 ) :

Options

  1. AExists and is equal to 1 2
  2. BDoes not exist
  3. CExists and is equal to 1
  4. DExists and is equal to - 1 2

Correct answer

A. Exists and is equal to 1 2

Step-by-step solution

If 2 + 3 i in one of the roots, then 2 - 3 i would be other. Since coefficients of the equation are real. Let γ be the third root, then product of roots →     α   β   γ   = 13 2 2 + 3 i   2 - 3 i   ⋅ γ = 13 2   4 + 9 ⋅ γ = 13 2 ⇒ γ = 1 2 The value of k will come if we put x = 1 2 in the equation 2 ⋅ 1 8 - 9 4 + k ⋅ 1 2 - 13 = 0 ⇒ k 2 = 15 ⇒ k = 30 ∴     Equation will become 2 x

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