If one zeroes of polynomial (a^2+9) x^2+13x+16a in reciprocal of the other find the value of a

If one zeroes of polynomial (a^2+9) x^2+13x+16a in reciprocal of the o

1.	If one zeroes of polynomial (a^2+9) x^2+13x+16a in reciprocal of the other find the value of a
Answer» According to the question,\xa0one of the zeros is the reciprocal of the other,\xa0so, let us consider\xa0one zero to be x.Therefore, the other zero will be 1 / x, and the product of zeros will be 1.For any\xa0polynomial\xa0of the form ax2+ bx + c = 0,Sum of\xa0zeros\xa0= - b / aProduct of zeroes = c / aUsing these results for the equation given in the question (a2\xa0+ 9)x2\xa0+ 13x + 6a, we getThe product of zeros will be c / a = 6a / (a2+ 9) = 1⇒\xa0a2+ 9 = 6a⇒\xa0a2- 6a + 9 = 0 [rearranging terms]⇒\xa0a2\xa0– 3a – 3a + 9 = 0 [splitting middle term]⇒ a (a - 3) -3 (a - 3) = 0 [taking a as common in 1st two terms and - 3 as common in last two terms]⇒ (a - 3) (a - 3) = 0\xa0⇒\xa0a = 3 , 3Hence, the value of a is 3.