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If one zeroes of polynomial (a^2+9) …

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

Posted by Kashif Alam 3 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Vishwas Kumar 3 years, 1 month ago

According to the question, one of the zeros is the reciprocal of the other, so, let us consider one zero to be x. Therefore, the other zero will be 1 / x, and the product of zeros will be 1. For any polynomial of the form ax2+ bx + c = 0, Sum of zeros = - b / a Product of zeroes = c / a Using these results for the equation given in the question (a2 + 9)x2 + 13x + 6a, we get The product of zeros will be c / a = 6a / (a2+ 9) = 1 ⇒ a2+ 9 = 6a ⇒ a2- 6a + 9 = 0 [rearranging terms] ⇒ a2 – 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 ⇒ a = 3 , 3 Hence, the value of a is 3.

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