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Let the first Zeros be a
and second be 1/a



a² + 9 = 6a
a² - 6a + 9 = 0
a² - 3a - 3a + 9 = 0
a ( a - 3 ) - 3 ( a - 3 ) = 0
( a - 3 ) ( a - 3 ) = 0
* ( a - 3 ) = 0
a = 3
* ( a - 3 ) = 0
a = 3
So , value of a is 3
and second be 1/a
a² + 9 = 6a
a² - 6a + 9 = 0
a² - 3a - 3a + 9 = 0
a ( a - 3 ) - 3 ( a - 3 ) = 0
( a - 3 ) ( a - 3 ) = 0
* ( a - 3 ) = 0
a = 3
* ( a - 3 ) = 0
a = 3
So , value of a is 3
Shanayarokz:
it is given in the question that the Zeros are reciprocal of each other!
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