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