solve 2(x²+1/x²)-(x+1/x)=11
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Answered by
35
Hi ,
It is given that b,
2( x² + 1/x² ) - ( x + 1/x ) = 11
2[ ( x + 1/x )² - 2 ] - ( x + 1/x ) - 11 = 0
Let x + 1/x = a ---( 1 )
2 ( a² - 2 ) - a - 11 = 0
2a² - 4 - a - 11 = 0
2a² - a - 15 = 0
2a² - 6a + 5a - 15 = 0
2a( a - 3 ) + 5 ( a - 3 ) = 0
( a - 3 )( 2a + 5 ) = 0
a - 3 = 0 or 2a + 5 = 0
now , from ( 1 ) ,
x + 1/x - 3 = 0 or 2 ( x + 1/x ) + 5 = 0
x² + 1 - 3x = 0 or 2x² + 2 + 5x = 0
x² - 3x + 1 = 0 or 2x² + 5x + 2 = 0
or 2x² + 4x + x + 2 = 0
or 2x( x + 2 ) + ( x + 2 ) = 0
or ( x + 2 ) ( 2x + 1 ) = 0
Therefore ,
x = -2 , x = -1/2 , x² -3x +1 = 0
I hope this helps you.
: )
It is given that b,
2( x² + 1/x² ) - ( x + 1/x ) = 11
2[ ( x + 1/x )² - 2 ] - ( x + 1/x ) - 11 = 0
Let x + 1/x = a ---( 1 )
2 ( a² - 2 ) - a - 11 = 0
2a² - 4 - a - 11 = 0
2a² - a - 15 = 0
2a² - 6a + 5a - 15 = 0
2a( a - 3 ) + 5 ( a - 3 ) = 0
( a - 3 )( 2a + 5 ) = 0
a - 3 = 0 or 2a + 5 = 0
now , from ( 1 ) ,
x + 1/x - 3 = 0 or 2 ( x + 1/x ) + 5 = 0
x² + 1 - 3x = 0 or 2x² + 2 + 5x = 0
x² - 3x + 1 = 0 or 2x² + 5x + 2 = 0
or 2x² + 4x + x + 2 = 0
or 2x( x + 2 ) + ( x + 2 ) = 0
or ( x + 2 ) ( 2x + 1 ) = 0
Therefore ,
x = -2 , x = -1/2 , x² -3x +1 = 0
I hope this helps you.
: )
Answered by
32
Step-by-step explanation:
or 2x² + 4x + x + 2 = 0
or 2x(x + 2) + (x + 2) = 0
or (x + 2) (2x + 1) = 0
Therefore,
x = -2, x = -1/2, x²-3x +1 = 0
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