Question

If ${{a}^{2}}-4a-1=0,a\ne 0$, then the value of ${{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}$ is,
(a) 24
(b) 26
(c) 28
(d) 30

Answer 1

Hint: We will solve the question by finding out the value of $a-\dfrac{1}{a}$ which will be carried out by ${{a}^{2}}-4a-1=0,a\ne 0$. After this we will find the value of $\left( a+\dfrac{1}{a} \right)$ by considering the equation like so ${{\left( a+\dfrac{1}{a} \right)}^{2}}={{\left( a-\dfrac{1}{a} \right)}^{2}}+4$. By doing all these steps we will substitute the values which are required in the solving the expression ${{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}$. This will help in solving the question properly.

Complete step-by-step answer:
As we are given ${{a}^{2}}-4a-1=0,a\ne 0$ then after simplifying it we will get
$\begin{align}
  & {{a}^{2}}-4a-1=0 \\
 & \Rightarrow {{a}^{2}}-4a=1 \\
\end{align}$
After dividing it by ‘a’ we will get
$\begin{align}
  & a-4=\dfrac{1}{a} \\
 & \Rightarrow a-\dfrac{1}{a}=4 \\
\end{align}$
Now to find the value of $\left( a+\dfrac{1}{a} \right)$. For this we are going to consider here the equation like so,
${{\left( a+\dfrac{1}{a} \right)}^{2}}={{\left( a-\dfrac{1}{a} \right)}^{2}}+4$. Now we will substitute the value of $a-\dfrac{1}{a}=4$ in the expression and this will result into
$\begin{align}
  & {{\left( a+\dfrac{1}{a} \right)}^{2}}={{\left( 4 \right)}^{2}}+4 \\
 & \Rightarrow {{\left( a+\dfrac{1}{a} \right)}^{2}}=16+4 \\
 & \Rightarrow {{\left( a+\dfrac{1}{a} \right)}^{2}}=20 \\
\end{align}$
Now we will consider the equation ${{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}$. This expression can be written also as ${{a}^{2}}+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}+3a$. And after taking 3 as common we will get the expression like so,${{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}={{a}^{2}}+\dfrac{1}{{{a}^{2}}}+3\left( a-\dfrac{1}{a} \right)$
As we have ${{\left( a+\dfrac{1}{a} \right)}^{2}}={{a}^{2}}+\dfrac{1}{{{a}^{2}}}+2a\left( \dfrac{1}{a} \right)$ by the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ so, we can also have ${{a}^{2}}+\dfrac{1}{{{a}^{2}}}={{\left( a+\dfrac{1}{a} \right)}^{2}}-2a\left( \dfrac{1}{a} \right)$. At this step we are going to substitute the value of ${{\left( a+\dfrac{1}{a} \right)}^{2}}=20$ in the equation. Thus, we get
$\begin{align}
  & {{a}^{2}}+\dfrac{1}{{{a}^{2}}}=\left( 20 \right)-2a\left( \dfrac{1}{a} \right) \\
 & \Rightarrow {{a}^{2}}+\dfrac{1}{{{a}^{2}}}=20-2 \\
 & \Rightarrow {{a}^{2}}+\dfrac{1}{{{a}^{2}}}=18 \\
\end{align}$
Now we will again consider ${{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}={{a}^{2}}+\dfrac{1}{{{a}^{2}}}+3\left( a-\dfrac{1}{a} \right)$ and substitute the obtained values of the above expressions. This result into
$\begin{align}
  & {{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}=18+3\left( 4 \right) \\
 & \Rightarrow {{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}=18+12 \\
 & \Rightarrow {{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}=30 \\
\end{align}$
Hence, the correct option is (d).

Note: Alternate method to find the value of ${{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}$ can also be carried out as by considering first the equation ${{a}^{2}}-4a-1=0,a\ne 0$. Now we will use the formula given by $a=\dfrac{-b\pm \sqrt{{{b}^{2}}-4{{a}_{1}}c}}{2{{a}_{1}}}$ where x are roots of quadratic equation of the form ${{a}_{1}}{{x}^{2}}+bx+c=0$ and $2{{a}_{1}}\ne 0$. By doing this we will get the value of ‘a’ and after that we will substitute the value in ${{a}^{2}}+3a+\dfrac{1}{{{a}^{2}}}-\dfrac{3}{a}$ to get the desired answer. There is no need to expand the equation ${{\left( a+\dfrac{1}{a} \right)}^{2}}={{\left( a-\dfrac{1}{a} \right)}^{2}}+4$ by using algebraic formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. Instead we will directly substitute the value of $a-\dfrac{1}{a}=4$ in the equation and we will solve further otherwise, the expansion will lead to nowhere.