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)​

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

Answers ( )

    0
    2021-08-18T08:55:02+00:00

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

    = > a² + b² = ±14/9

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )