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If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is reciprocal of the other, find the value of ‘a’. |
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Answer» Let one zero of the given polynomial is α. Given that one zero of given polynomial is reciprocal of the other. ∴ The other zero of given polynomial is \(\frac{1}{a}.\) Hence, the zeros of the given polynomial (α2 + 9)x2 + 13x + 6 are and \(\frac{1}{a}.\) Now, the product of the zeros is α. \(\frac{1}{a}\) = \(\frac{6a}{a^2+9}\) ⇒ \(\frac{6a}{a^2+9}=1\) ⇒ α2 + 9 = 6a ⇒ α2 − 6 + 9 = 0 ⇒ (α − 3)2 = 0 (∵ (α − b)2 = α2 − 2ab + b2) ⇒ α = 3. Hence, the value of is 3. |
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