Question

If we have an expression as $\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}+\dfrac{4+\sqrt{5}}{\sqrt{5}+4}=a\sqrt{3}+b\sqrt{5}+c$ then find the value of $a+b+c$.

Answer 1

Hint: In this problem we have the two irrational numbers one is $\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}$ and $\dfrac{4+\sqrt{5}}{\sqrt{5}+4}$. We need to rationalize the two terms to proceed further. We know that if $\dfrac{1}{\sqrt{a}+\sqrt{b}}$ is irrational number to rationalize it we need to multiply with $\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}$. By using the above method, we will rationalize the both terms $\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}$ and $\dfrac{4+\sqrt{5}}{\sqrt{5}+4}$ . Further we will simplify the obtained equation and equate it to the given value to find the values of $a$, $b$, $c$. After getting the values of $a$, $b$, $c$ we can find required value i.e. $a+b+c$.

Complete step-by-step solution:
Given that,
$\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}+\dfrac{4+\sqrt{5}}{\sqrt{5}+4}=a\sqrt{3}+b\sqrt{5}+c$
Consider the term $\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}$. To rationalize we will multiply $\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}$ with $\dfrac{2\sqrt{3}+2}{2\sqrt{3}+2}$, then we will get
$\begin{align}
  & \dfrac{2+\sqrt{3}}{2\sqrt{3}-2}=\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}\times \dfrac{2\sqrt{3}+2}{2\sqrt{3}+2} \\
 & \Rightarrow \dfrac{2+\sqrt{3}}{2\sqrt{3}-2}=\dfrac{\left( 2+\sqrt{3} \right)\left( 2\sqrt{3}+2 \right)}{\left( 2\sqrt{3}-2 \right)\left( 2\sqrt{3}+2 \right)} \\
\end{align}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, then we will get
$\begin{align}
  & \Rightarrow \dfrac{2+\sqrt{3}}{2\sqrt{3}-2}=\dfrac{2\left( 2\sqrt{3}+2 \right)+\sqrt{3}\left( 2\sqrt{3}+2 \right)}{{{\left( 2\sqrt{3} \right)}^{2}}-{{2}^{2}}} \\
 & \Rightarrow \dfrac{2+\sqrt{3}}{2\sqrt{3}-2}=\dfrac{4\sqrt{3}+4+2\left( 3 \right)+2\sqrt{3}}{4\left( 3 \right)-4} \\
 & \Rightarrow \dfrac{2+\sqrt{3}}{2\sqrt{3}-2}=\dfrac{10+6\sqrt{3}}{8}....\left( \text{i} \right) \\
\end{align}$
Consider the term $\dfrac{4+\sqrt{5}}{\sqrt{5}+4}$. To rationalize we will multiply $\dfrac{4+\sqrt{5}}{\sqrt{5}+4}$ with $\dfrac{\sqrt{5}-4}{\sqrt{5}-4}$, then we will get
$\begin{align}
  & \dfrac{4+\sqrt{5}}{\sqrt{5}+4}=\dfrac{4+\sqrt{5}}{\sqrt{5}+4}\times \dfrac{\sqrt{5}-4}{\sqrt{5}-4} \\
 & \Rightarrow \dfrac{4+\sqrt{5}}{\sqrt{5}+4}=\dfrac{\left( 4+\sqrt{5} \right)\left( \sqrt{5}-4 \right)}{\left( \sqrt{5}+4 \right)\left( \sqrt{5}-4 \right)} \\
\end{align}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, then we will get
$\begin{align}
  & \Rightarrow \dfrac{4+\sqrt{5}}{\sqrt{5}+4}=\dfrac{4\left( \sqrt{5}-4 \right)+\sqrt{5}\left( \sqrt{5}-4 \right)}{{{\left( \sqrt{5} \right)}^{2}}-{{4}^{2}}} \\
 & \Rightarrow \dfrac{4+\sqrt{5}}{\sqrt{5}+4}=\dfrac{4\sqrt{5}-16+5-4\sqrt{5}}{5-16} \\
 & \Rightarrow \dfrac{4+\sqrt{5}}{\sqrt{5}+4}=1....\left( \text{ii} \right) \\
\end{align}$
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ we have the values of $\dfrac{2+\sqrt{3}}{2\sqrt{3}-2}$ and $\dfrac{4+\sqrt{5}}{\sqrt{5}+4}$. So, substituting these values in the given equation, then we will get
$\begin{align}
  & \dfrac{2+\sqrt{3}}{2\sqrt{3}-2}+\dfrac{4+\sqrt{5}}{\sqrt{5}+4}=a\sqrt{3}+b\sqrt{5}+c \\
 & \Rightarrow \dfrac{10+6\sqrt{3}}{8}+1=a\sqrt{3}+b\sqrt{5}+c \\
 & \Rightarrow \left( \dfrac{10}{8} \right)+\left( \dfrac{6\sqrt{3}}{8} \right)+1=a\sqrt{3}+b\sqrt{5}+c \\
 & \Rightarrow \left( \dfrac{3}{4} \right)\sqrt{3}+\dfrac{5}{4}+1=a\sqrt{3}+b\sqrt{5}+c \\
 & \Rightarrow \left( \dfrac{3}{4} \right)\sqrt{3}+\left( 0 \right)\sqrt{5}+\left( \dfrac{9}{4} \right)=a\sqrt{3}+b\sqrt{5}+c \\
\end{align}$
Equating the above equation we will get the values of $a$, $b$, $c$ as $a=\dfrac{3}{4}$, $b=0$, $c=\dfrac{9}{4}$. Now the value of $a+b+c$ is given by
$\begin{align}
  & a+b+c=\dfrac{3}{4}+0+\dfrac{9}{4} \\
 & \Rightarrow a+b+c=\dfrac{3+9}{4} \\
 & \Rightarrow a+b+c=3 \\
\end{align}$
$\therefore $ The value of $a+b+c$ is $3$.

Note: Here we can observe that denominator and the numerator in the term $\dfrac{4+\sqrt{5}}{\sqrt{5}+4}$ are equal so that you can cancel them and write the value of the term $\dfrac{4+\sqrt{5}}{\sqrt{5}+4}$ as $1$. But it is not recommended because the term is irrational and we need to rationalize it.