NTA Abhyas JEE Main2020MathematicsQuadratic EquationPractice
Let α and β be the roots of the equation x 2 + ax + 1 = 0 , a ≠ 0 . Then the equation whose roots are - α + 1 β and - 1 α + β is
Options
- Ax 2 = 0
- Bx 2 + 2 a x + 4 = 0
- Cx 2 - 2 a x + 4 = 0
- Dx 2 - a x + 1 = 0
Correct answer
C. x 2 - 2 a x + 4 = 0
Step-by-step solution
α + β = - a and α β = 1 Let S and P be the sum and product of the roots of the required equation. Then, S = - α - 1 β - 1 α - β = - ( α + β ) - 1 α + 1 β = - ( α + β ) - α + β α β = - ( - a ) - - a 1 = 2 a P = - α + 1 β - 1 α + β = 1 + α β + 1 α β + 1 = 1 + 1 + 1 + 1 = 4 So, the required equation is x 2 - S x + P = 0 i.e. x 2 - 2 a x + 4 = 0