Algebra

Problem : 19 )
if x + y = 1, and (x² + y²) (x³ + y³ ) = 26, then 
what is the value of (x² + y²) ?


Solution : 


Problem : Given x + y = 1 and (x² + y²) (x³ + y³ ) = 26

. .(x² + y²) (x³ + y³ ) = 26
=> [(x + y)² - 2xy] [(x + y)³ - 3xy (x + y)] = 26
=> (1 - 2xy) (1 - 3xy) = 26

=> 1 - 5xy + 6x²y² = 26
=> 6x²y² - 5xy - 25 = 0
=> 6x²y² - 15xy + 10xy - 25 = 0

=> (3xy + 5)(2xy - 5) = 0
=> xy = -5/3 and xy = 5/2

Now, x² + y² = (x + y)² - 2xy
=> x² + y² = 1 - 2xy
=> x² + y² = 1 - 2 * (-5/3)

=> 13 / 3 ==> (Answer)

Note : If we take xy = 5/2, x² + y² = -ve , which is not possible as sum of two real squares will never be negative