Answer

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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.

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.

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