find the zeroes of the quadratic polynomial x^2+5x+6 then find the value of 1/a^2+1/B^2
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Hi ,
Given quadratic equation ,
p( x ) = x² + 5x + 6
compare p( x ) with Ax² + Bx + C , we get
A = 1 , B = 5 , C = 6
and it is given that a , b are zeroes of p ( x ).
1 ) sum of the zeroes = -B/A
a + b = - 5/1 ---( 1 )
2 ) product of the zeroes = C/A
ab = 6 ---( 2 )
Now ,
1/a² + 1/b² = ( b² + a² )/( ab )²
= [ ( a + b )² - 2ab ]/( ab )²
= [ ( -5 )² - 2 × 6 ] / 6²
= [ 25 - 12 ] / 36
= 13/36
I hope this helps you.
: )
Given quadratic equation ,
p( x ) = x² + 5x + 6
compare p( x ) with Ax² + Bx + C , we get
A = 1 , B = 5 , C = 6
and it is given that a , b are zeroes of p ( x ).
1 ) sum of the zeroes = -B/A
a + b = - 5/1 ---( 1 )
2 ) product of the zeroes = C/A
ab = 6 ---( 2 )
Now ,
1/a² + 1/b² = ( b² + a² )/( ab )²
= [ ( a + b )² - 2ab ]/( ab )²
= [ ( -5 )² - 2 × 6 ] / 6²
= [ 25 - 12 ] / 36
= 13/36
I hope this helps you.
: )
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