If one zero of the polynomial $\left( {{a^2} + 9} \right){x^2} + 13x + 6a$ is the reciprocal of the other, then find the value of $a$.
Answer
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Hint: In this question, we are given a polynomial and we have been told the relation between its two zeroes. Using that relation, we have to find the value of $a$. Start by assuming two zeroes of the given polynomial. Write one zero in terms of the other using the given relation. Multiply both the zeroes with each other and write the product equal to the ratio of $c$ and $a$. If you shift the denominator to the other side, you will get a quadratic equation. Solve the equation and you will have the required answer.
Formula used: Product of zeroes = $\dfrac{c}{a}$.
Complete step-by-step solution:
We are given a polynomial $\left( {{a^2} + 9} \right){x^2} + 13x + 6a$. If we compare the given polynomial with $a{x^2} + bx + c = 0$, we will get the following values:
$ \Rightarrow a = {a^2} + 9$, $b = 13$ and $c = 6a$
Let us assume that the given polynomial has two zeroes: $\alpha $ and $\beta $.
We know that the product of the zeroes = $\dfrac{c}{a}$.
Using this relation to find the value of $a$.
$ \Rightarrow \alpha \beta = \dfrac{{6a}}{{{a^2} - 9}}$
But we are also given that one zero is the reciprocal of the other. Therefore, we will substitute $\beta = \dfrac{1}{\alpha }$.
$ \Rightarrow \alpha \dfrac{1}{\alpha } = \dfrac{{6a}}{{{a^2} - 9}}$
Simplifying,
$ \Rightarrow 1 = \dfrac{{6a}}{{{a^2} - 9}}$
Shifting denominator and numerator to the other side –
$ \Rightarrow {a^2} - 6a + 9 = 0$
Now, we will find the value of $a$ using splitting the middle term,
$ \Rightarrow {a^2} - 3a - 3a + 9 = 0$
Making factors,
$ \Rightarrow a\left( {a - 3} \right) - 3\left( {a - 3} \right) = 0$
Simplifying we get,
$ \Rightarrow \left( {a - 3} \right)\left( {a - 3} \right) = 0$
Hence,
$ \Rightarrow a = 3$
$\therefore $ The required value of $a = 3$
Note: There is another property which says that the sum of zeroes of a polynomial $ = \dfrac{{ - b}}{a}$. But we will not use this property here. It is because when we use these properties, our aim is to simplify the equation in such a way that we can find the required value easily. If we use the sum of zeroes property, it will give us an equation in two variables which will make it impossible to solve. Hence, that is why we used the product of zeroes property.
Formula used: Product of zeroes = $\dfrac{c}{a}$.
Complete step-by-step solution:
We are given a polynomial $\left( {{a^2} + 9} \right){x^2} + 13x + 6a$. If we compare the given polynomial with $a{x^2} + bx + c = 0$, we will get the following values:
$ \Rightarrow a = {a^2} + 9$, $b = 13$ and $c = 6a$
Let us assume that the given polynomial has two zeroes: $\alpha $ and $\beta $.
We know that the product of the zeroes = $\dfrac{c}{a}$.
Using this relation to find the value of $a$.
$ \Rightarrow \alpha \beta = \dfrac{{6a}}{{{a^2} - 9}}$
But we are also given that one zero is the reciprocal of the other. Therefore, we will substitute $\beta = \dfrac{1}{\alpha }$.
$ \Rightarrow \alpha \dfrac{1}{\alpha } = \dfrac{{6a}}{{{a^2} - 9}}$
Simplifying,
$ \Rightarrow 1 = \dfrac{{6a}}{{{a^2} - 9}}$
Shifting denominator and numerator to the other side –
$ \Rightarrow {a^2} - 6a + 9 = 0$
Now, we will find the value of $a$ using splitting the middle term,
$ \Rightarrow {a^2} - 3a - 3a + 9 = 0$
Making factors,
$ \Rightarrow a\left( {a - 3} \right) - 3\left( {a - 3} \right) = 0$
Simplifying we get,
$ \Rightarrow \left( {a - 3} \right)\left( {a - 3} \right) = 0$
Hence,
$ \Rightarrow a = 3$
$\therefore $ The required value of $a = 3$
Note: There is another property which says that the sum of zeroes of a polynomial $ = \dfrac{{ - b}}{a}$. But we will not use this property here. It is because when we use these properties, our aim is to simplify the equation in such a way that we can find the required value easily. If we use the sum of zeroes property, it will give us an equation in two variables which will make it impossible to solve. Hence, that is why we used the product of zeroes property.
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