Question

if zeroes of the polynomial p (x)=(a^2+9)x^2 +45x+6a is reciprocal of other. find the value of a.

Answer 1

Hi there!

♦ Given :-
Quadratic poly. is p(x) = (a² + 9)x² + 45x + 6a

Let the zeroes be k and (1 / k)
We know,

The sum of zeroes (k + 1 / k) = –45 / (a² + 9)

Product of zeroes [k × (1 / k)] = 6a / (a² + 9)

⇒ 6a / (a² + 9) = 1

⇒ (a² + 9) = 6a

⇒ a² – 6a + 9 = 0

⇒ a² – 2(a)(3) + 32 = 0

⇒ (a – 3)² = 0

⇒ (a – 3) = 0

∴ a = 3

Hence, The required answer is :-
The value of a = 3

hope it helps! :)

Answer 2

Hiii friend,


Let the one zero of the given polynomial be Alpha .

Then the other zero = 1/Alpha


P(X) = (A²+9)X²+45X+6A


Here,


A = (A²+9) ‍, B = 45 and C = 6A


Product of zeroes = C/A


Alpha × 1/Alpha = 6A/A²+9

1 = 6A/A²+9


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



Hence,


The Value of A is 3



HOPE IT WILL HELP YOU...... :-)