if x=√3+1/√3-1 y=√3-1/√3+1 find the value of x sq-xy+y sq
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x = 2 + √3 , plz check
y = 2 - √3
my answer is right but your answer is wrong.... the answer should be 16
bro please mark my answer as brainliest and start following me
wrong answer
how???
I already posted above x , y values
you are wrong
thanks for marking my answer as brainliest ☺️
Answered by
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Hi ,
x = ( √3 + 1 ) / ( √3 - 1 )
y = ( √3 - 1 ) / ( √3 + 1 )
x + y
= [(√3+1)/(√3-1 )] + [ (√3-1 )/(√3 + 1 )]
= [ (√3 + 1)² + (√3 -1 )²]/[(√3)² - 1 ]
= { 2[( √3 )² + 1 ]}/ 2
= 3 + 1
= 4
x + y = 4 ----( 1 )
xy = 1 ----( 2 )
Now ,
x² - xy + y²
= ( x + y )² -3xy
= 4² - 3 × 1
= 16 - 3
= 13
I hope this helps you.
: )
x = ( √3 + 1 ) / ( √3 - 1 )
y = ( √3 - 1 ) / ( √3 + 1 )
x + y
= [(√3+1)/(√3-1 )] + [ (√3-1 )/(√3 + 1 )]
= [ (√3 + 1)² + (√3 -1 )²]/[(√3)² - 1 ]
= { 2[( √3 )² + 1 ]}/ 2
= 3 + 1
= 4
x + y = 4 ----( 1 )
xy = 1 ----( 2 )
Now ,
x² - xy + y²
= ( x + y )² -3xy
= 4² - 3 × 1
= 16 - 3
= 13
I hope this helps you.
: )
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