if one zero of (a^4+4)x^2+13x+4a is reciprocal of other. find a
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Hi ,
( plz check the problem again , constant
term may be 4a^2 )
let p( x ) = ( a^4 + 4 )x^2 + 13x + 4a^2
according to the problem
one zero is reciprocal to another.
let
one zero = p
second zero = 1/ p
product of the zeroes
= constant / ( x^2 coefficient )
p × 1/ p = ( 4a^2 )/ ( a^4 + 4 )
1 = ( 4a^2 )/ ( a^4 + 4 )
a^4 + 4 = 4a^2
a^4 - 4a^2 + 4 = 0
( a^2 )^2 - 2 × a^2 × 2 + 2^2 = 0
(a^2 - 2 )^2 = 0
Therefore,
a^2 - 2 = 0
a^2 = 2
a = + or - (sqrt 2)
i hope this will useful to you.
****
( plz check the problem again , constant
term may be 4a^2 )
let p( x ) = ( a^4 + 4 )x^2 + 13x + 4a^2
according to the problem
one zero is reciprocal to another.
let
one zero = p
second zero = 1/ p
product of the zeroes
= constant / ( x^2 coefficient )
p × 1/ p = ( 4a^2 )/ ( a^4 + 4 )
1 = ( 4a^2 )/ ( a^4 + 4 )
a^4 + 4 = 4a^2
a^4 - 4a^2 + 4 = 0
( a^2 )^2 - 2 × a^2 × 2 + 2^2 = 0
(a^2 - 2 )^2 = 0
Therefore,
a^2 - 2 = 0
a^2 = 2
a = + or - (sqrt 2)
i hope this will useful to you.
****
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