Question

Guys plzz answer it ....... :)

if one zero of the polynomial (a square + 9) x square + 13 x + 6a is reciprocal of the Other. find the value of a.

Answer 1

HELLO DEAR,

. LET ONE OF THE IT ZEROS BE P
AMD THE OTHER BE 1/P


NOE WE KNOW THAT ,



PRODUCT OF THE ROOTS = P x 1/P
=> (CONSTANT TERM) / coefficient Of X²


⇒ (6a)/(a2 + 9) = p x 1/p

                             ⇒ (6a) / (a2 + 9) = 1
                 
                             ⇒ a2 - 6a + 9 = 0

                             ⇒ (a - 3)2 = 0

                             ⇒ a - 3 = 0

                             ⇒ a = 3

THEREFORE, THE VALUE OF A IS 3.


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Solution:

p(x) = (a²+9)x²+13x+6a

Let the one zero of given polynomial be k

then the other zero will be 1/k

Product of roots = C/A

k×(1/k) = 6a/a²+9

1 = 6a/a²+9

a²+9=6a

a²-6a+9 = 0

(a-3)² = 0

=> a-3 = 0

=> a = 3. Ans.