Question

3a²=4a-5 and 3b²=4b-5 where a≠b then find the value of (a²+b²)​

Answer 1

Answer:

±14/9

Step-by-step explanation:

= > 3a² - 3b² = ( 4a - 5 ) - ( 4b - 5 )

= > 3( a² - b² ) = 4a - 5 - 4b + 5

= > 3( a + b )( a - b ) = 4a - 4b

= > 3( a + b )( a - b ) = 4( a - b )

= > 3( a + b ) = 4

= > a + b = 4/3 ... (1)

Multiply both -

= > 3a² *3b² = ( 4a - 5 )( 4b - 5 )

= > 9a²b² = 4a*4b - 4a*5 - 5*4b + 25

= > 9a²b² = 16ab - 20a - 20b + 25

= > 9(ab)² = 16ab - 20( a + b ) + 25

= > 9(ab)² = 16ab - 20(4/3) + 25

Let ab = x

= > 9x² = 16x - (80/3) + 25

= > 27x² = 48x - 80 + 75

= > 27x² - 48x - 5 = 0

= > ( 9x - 1 )( 3x - 5 ) = 0

= > x = 1/9 or x = 5/3

= > ab = 1/9 or ab = 5/3 ... (2)

Square on both sides of (1):

= > (a+b)² = (4/3)²

= > a² + b² + 2ab = 16/9

= > a² + b² = 16/9 - 2ab

From (2)

= > a² + b² = 16/9 - 2(1/9) or 16/9 - 2(5/3)

= > a² + b² = (16-2)/9 or (16-30)/9

= > a² + b² = 14/9 or - 14/9

Square on any real number can't be negative, so a² + b² should be positive, but here equations may have imaginary roots. So,

= > a² + b² = ±14/9