Question

If one zero of the polynomial \[({{a}^{2}}-9){{x}^{2}}+13x+6a\] is the reciprocal of the other, find the value of ‘a’.

Answer 1

Hint: If the coefficient of \[{{x}^{2}}\] is 1, then we can represent the quadratic equation as follows
\[{{x}^{2}}-bx+c=0\]
(Where the value of b is the sum of the two roots of the quadratic equation and the value of c is the product of the two roots of the quadratic equation)
Now, to convert the given quadratic equation into the above mentioned form, we can just divide the equation with the coefficient of \[{{x}^{2}}\] and we have to make the coefficient of x negative as well.

Complete step-by-step answer:
As mentioned in the question, we have to find the value of a.
Now, as mentioned in the hint, we can convert the given quadratic equation into the mentioned form by just dividing the equation with the coefficient of \[{{x}^{2}}\] and also making the coefficient of x negative as follows
\[{{x}^{2}}-\dfrac{\left( -13 \right)}{({{a}^{2}}-9)}x+\dfrac{6}{({{a}^{2}}-9)}a=0\]
(The condition which is required to do the above transformation is that the value of \[({{a}^{2}}-9)\] cannot be zero)
Now, as the roots of this equation are reciprocal of each other, therefore, the product of the roots will be 1. Hence, as we know that the constant part of the equation is equal to the product of the roots of this equation, we can write the following
\[\begin{align}
 \Rightarrow & 1=\dfrac{6a}{({{a}^{2}}-9)} \\
 \Rightarrow & ({{a}^{2}}-9)=6a \\
 \Rightarrow & {{a}^{2}}-6a-9=0 \\
\end{align}\]
We know that the roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] are
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\begin{align}
 \Rightarrow & a=\dfrac{+6\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times -9}}{2\times 1} \\
\Rightarrow & a=\dfrac{6\pm \sqrt{36+36}}{2\times 1} \\
 \Rightarrow & a=\dfrac{6\pm 6\sqrt{2}}{2} \\
\Rightarrow & a=7.243,-1.243 \\
\end{align}\]
Hence, the values of a are 7.243 and -1.243.

Note: The students can make an error if they don’t know how to solve a quadratic equation or how to get the roots of a quadratic equation that is as follows
The roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] are
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] .
We can solve quadratic equations using the factorisation method as well.