71.

The roots α and β of a quadratic equation, satisfy the relations α + β = α² + β² and αβ = α²β². What is the number of such quadratic equations ?

• 0

• 2

• 3

• 4

Consider following two relations : 

α + β = α² + β²   ...(1)

and

αβ = α²β²             ...(2)

First of all we will find values of α and β  using hit and trial such that both above mentioned relations are satisfied.

Case 1: α = 1, β = 0

Sum of roots = 1

Product of roots = 0

Equation :

x² - (Sum of roots)x + (Product of roots) = 0

$\Rightarrow {\mathrm{x}}^2 - \mathrm{x} = 0$

Case 2 :  α = 0, β = 0

Sum of roots = 0

Product of roots = 0

Equation : 

x² - (Sum of roots)x + (Product of roots) = 0

$\Rightarrow {\mathrm{x}}^2 = 0$

Case 3 : 

 α = 1, β = 1

Sum of roots = 1

Product of roots = 1

Equation : 

x² - (Sum of roots)x + (Product of roots) = 0

$\Rightarrow {\mathrm{x}}^2 - 2\mathrm{x} + 1 = 0$

Case 4 : 

$\mathrm{\alpha } = \mathrm{\omega }, \mathrm{\beta } = {\mathrm{\omega }}^2$

Sum of roots = $\mathrm{\omega } + {\mathrm{\omega }}^2$ = - 1

Product of roots = $\mathrm{\omega } . {\mathrm{\omega }}^2$ = ${\mathrm{\omega }}^3 = 1$

Equation : 

x² - (Sum of roots)x + (Product of roots) = 0

$\Rightarrow {\mathrm{x}}^2 - \mathrm{x} + 1 = 0$

So total number of possible equation = 4