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Last updated date: 28th Nov 2023
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# If one zero of the polynomials $\left( {{a^2} + 9} \right){x^2} + 13x + 6a$ is the reciprocal of the other, then the value of $a$ is

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Hint: In the given polynomial $\left( {{a^2} + 9} \right){x^2} + 13x + 6a$ the degree (highest power) is two. So, the polynomial has two zeros or roots.

Let $p$ be one of the zero of the polynomials $\left( {{a^2} + 9} \right){x^2} + 13x + 6a$. Then the other zero is $\dfrac{1}{p}$ since one of the zero is the reciprocal of the other.
We know that for the quadratic polynomial $a{x^2} + bx + c = 0$, the sum of the roots is $- \dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$.
Now consider the product of the zeroes or roots
i.e. $p \times \dfrac{1}{p} = \dfrac{{6a}}{{{a^2} + 9}}$
$1 = \dfrac{{6a}}{{{a^2} + 9}} \\ {a^2} + 9 = 6a \\ {a^2} - 6a + 9 = 0 \\$
By solving ${a^2} - 6a + 9 = 0$ we get
${a^2} - 3a - 3a + 9 = 0 \\ a\left( {a - 3} \right) - 3\left( {a - 3} \right) = 0 \\ \left( {a - 3} \right)\left( {a - 3} \right) = 0 \\ {\left( {a - 3} \right)^2} = 0 \\ \therefore a = 3 \\$
Therefore, the value of $a$ is $3$.

Note: The given polynomial is quadratic polynomial since the degree (highest power) is two. In this problem we can also consider the sum of the zeroes or roots but it is a lengthy process. So, we considered the product of the zeroes or roots.