If the zeroes of the polynomial f(x) = x3 – 12x2 + 39x + a are in AP, find the value of a.
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Answered by
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Hey there
let ( x -y ) , x , ( x + y ) are the zeroes of the polynomial which are in AP
According to the Question :-
x - y + x + x + y = 12 / 1
⇒ 3x = 12
⇒ x = 4
Now the other Zeroes will be -
4 , ( 4 + y ) , ( 4 - y )
If we putt 4 in the f ( x ) = 0 we can easily get the value of a
therefore ,
f ( 4 ) = 0
⇒ x ³ - 12 x² + 39x + a = 0
⇒ ( 4 ) ³ - 12 ( 4 )² + 39 ( 4 ) + a = 0
⇒ 64 - 12 × 16 + 156 + a = 0
⇒64 - 192 + 156 + a = 0
⇒220 -192 + a = 0
⇒28 + a = 0
⇒ a = - 28
Hope this helps ya ☺
let ( x -y ) , x , ( x + y ) are the zeroes of the polynomial which are in AP
According to the Question :-
x - y + x + x + y = 12 / 1
⇒ 3x = 12
⇒ x = 4
Now the other Zeroes will be -
4 , ( 4 + y ) , ( 4 - y )
If we putt 4 in the f ( x ) = 0 we can easily get the value of a
therefore ,
f ( 4 ) = 0
⇒ x ³ - 12 x² + 39x + a = 0
⇒ ( 4 ) ³ - 12 ( 4 )² + 39 ( 4 ) + a = 0
⇒ 64 - 12 × 16 + 156 + a = 0
⇒64 - 192 + 156 + a = 0
⇒220 -192 + a = 0
⇒28 + a = 0
⇒ a = - 28
Hope this helps ya ☺
TheAishtonsageAlvie:
:)
Why u deleted my answer???
i didn't bro
Answered by
8
Answer:
let ( x -y ) , x , ( x + y ) are the zeroes of the polynomial which are in AP
According to the Question :-
x - y + x + x + y = 12 / 1
⇒ 3x = 12
⇒ x = 4
Now the other Zeroes will be -
4 , ( 4 + y ) , ( 4 - y )
If we putt 4 in the f ( x ) = 0 we can easily get the value of a
therefore ,
f ( 4 ) = 0
⇒ x ³ - 12 x² + 39x + a = 0
⇒ ( 4 ) ³ - 12 ( 4 )² + 39 ( 4 ) + a = 0
⇒ 64 - 12 × 16 + 156 + a = 0
⇒64 - 192 + 156 + a = 0
⇒220 -192 + a = 0
⇒28 + a = 0
⇒ a = - 28
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