Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202429 Jan 2024Morning ShiftMathematicsComplex NumberActual

Let α , β be the roots of the equation x 2 - x + 2 = 0 with Im ( α ) > Im ( β ) . Then α 6 + α 4 + β 4 - 5 α 2 is equal to

Correct answer

0

Step-by-step solution

Given equation x 2 - x + 2 has roots α & β , So, α + β = 1 , α β = 2 ⇒ α 4 + β 4 = α 2 + β 2 2 - 2 α 2 β 2 ⇒ α 4 + β 4 = α + β 2 - 2 α β 2 - 2 α 2 β 2 ⇒ α 4 + β 4 = 1 - 4 2 - 2 × 4 ⇒ α 4 + β 4 = 1 It is given that, α is a root of x 2 - x + 2 . ⇒ α 2 - α + 2 = 0 ⇒ α 2 = α - 2 ⇒ α 4 = α 2 + 4 - 4 α ⇒ α 4 = α - 2 + 4 - 4 α ⇒ α 4 = 2 - 3 α Now, α 6 - 5 α 2 = α 2 α 4 - 5 ⇒ α 6 - 5 α 2 = α - 2 2 - 3 α - 5 ⇒ α 6 - 5 α 2 = α - 2 - 3 α - 3 ⇒ α 6 - 5 α 2 = - 3 α 2 - α - 2 ⇒ α 6 - 5 α 2 = - 3 α - 2 - α - 2 ⇒ α 6 - 5 α 2 = 12

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