1.

(3-√2)^x+(3+√2)^=6 how to solve this answer is 1

Answer»

Answer:

1

Step-by-step explanation:

⇒ 3 + √2

⇒ ( 3 + √2 )( 3 - √2 ) / ( 3 - √2 )

     USING ( a + b )( a - b ) = a^2 - b^2

⇒ [ 3^2 - ( √2 )^2 ] / ( 3 - √2 )

⇒ [ 9 - 2 ] / ( 3 - √2 )

⇒ 7 / ( 3 - √2 )

    in the question, REPLACING ( 3 + √2 )^x by [ 7 / ( 3 - √2 ) ]^x

⇒ ( 3 - √2 )^x + [ 7 / ( 3 - √2 ) ]^x = 6

  Let ( 3 - √2 )^x = a

⇒ a + 7/a = 6

⇒ ( a^2 + 7 ) / a = 6

⇒ a^2 + 7 = 6a

⇒ a^2 - 6a + 7 = 0

    Using QUADRATIC eq.

⇒ a = [ 6 ± √( 36 - 28 ) / 2

⇒ a = [ 6 ± √8 ] / 2

⇒ a = [ 6 ± 2√2 ] / 2

⇒ a = 2( 3 ± 2√2 ) / 2

⇒ a = 3 ± √2

taking -ve value   HENCE,

⇒ a = ( 3 - √2 )^x  = 3 - √2

⇒ ( 3 - √2 )^x = ( 3 - √2 )^1

    Comparing

⇒ x = 1



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