Question

If one zeroes of the polynomial \[\left( {{a^2} + 9} \right){x^2} + 13x + 6a\] is reciprocal of the other. Find $a$.

Answer 1

Hint: We will first let the zero of the polynomial as $p$ and then according to the given condition, $\dfrac{1}{p}$ will also be the root of the given equation. Next, we know that if the equation is \[A{x^2} + Bx + C\], then the sum of the roots is given as $ - \dfrac{B}{A}$ and the product of the roots is given as $\dfrac{C}{A}$. Hence form an equation using the condition of the product of roots and solve it to find the value of $a$

Complete step-by-step answer:
Let the zero of the given polynomial \[\left( {{a^2} + 9} \right){x^2} + 13x + 6a\] is $p$, then according to the given condition $\dfrac{1}{p}$ is also the zero of the polynomial.
Now, we know that if we have the quadratic equation as\[A{x^2} + Bx + C\], then the sum of the roots is given as $ - \dfrac{B}{A}$ and the product of the roots is given as $\dfrac{C}{A}$
From, we will use the condition of the product of roots to form an equation.
If $p$ and $\dfrac{1}{p}$ are roots of \[\left( {{a^2} + 9} \right){x^2} + 13x + 6a\], then,
$
  \dfrac{{6a}}{{{a^2} + 9}} = p\left( {\dfrac{1}{p}} \right) \\
   \Rightarrow \dfrac{{6a}}{{{a^2} + 9}} = 1 \\
$
On cross multiplying we will get the expression as,
${a^2} - 6a + 9 = 0$
We will now factorise the equation.
$
  {a^2} - 3a - 3a + 9 = 0 \\
   \Rightarrow a\left( {a - 3} \right) - 3\left( {a - 3} \right) = 0 \\
   \Rightarrow \left( {a - 3} \right)\left( {a - 3} \right) = 0 \\
   \Rightarrow a = 3 \\
 $
So the value of a=3.

Note: We have used the condition that the product of roots of the equation is $\dfrac{C}{A}$ when the equation is \[A{x^2} + Bx + C\]. We cannot use the condition of the sum of roots as it will involve the expression $p + \dfrac{1}{p}$ which will give a complicated expression and moreover, there will be an additional variable$p$. Whereas, when we multiply $p$ and $\dfrac{1}{p}$ we are just left with 1.