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

Question

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

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4 weeks 2021-08-18T08:53:44+00:00 1 Answer 0 views 0

±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 + should be positive, but here equations may have imaginary roots. So,

= > a² + b² = ±14/9