Question

Add the following numbers\[\]
(i)$\left( 2\sqrt{3}-5\sqrt{2} \right)$ and $\left( \sqrt{3}+2\sqrt{2} \right)$\[\]
(ii)$\left( 2\sqrt{2}+5\sqrt{3}-7\sqrt{5} \right)$ and $\left( 3\sqrt{3}-\sqrt{2}+\sqrt{5} \right)$\[\]
(iii) $\left( \dfrac{2}{3}\sqrt{7}-\dfrac{1}{2}\sqrt{2}+6\sqrt{11} \right)$ and $\left( \dfrac{1}{3}\sqrt{7}+\dfrac{3}{2}\sqrt{2}-\sqrt{11} \right)$\[\]

Answer 1

Hint: The numbers given in the equation are irrational numbers. We open the brackets and write the terms with the same type of rational number, say $\sqrt{a}$ close to each other. We take $\sqrt{a}$ common and add up their multipliers in the brackets. We follow the rule of addition of fraction if the multipliers are fractions.\[\]

Complete step-by-step solution
We know that an irrational number is real number cannot be expressed in the form $\dfrac{p}{q},q\ne 0$.Some example of irrational numbers are square roots of prime for example $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{11}$ etc.\[\]
(i) We are given in the question the irrational numbers to add are$\left( 2\sqrt{3}-5\sqrt{2} \right)$ and $\left( \sqrt{3}+2\sqrt{2} \right)$. Let us proceed by writing the terms first and open the bracket following the BODMAS rule.
\[\begin{align}
  & \left( 2\sqrt{3}-5\sqrt{2} \right)+\left( \sqrt{3}+2\sqrt{2} \right) \\
 & =2\sqrt{3}-5\sqrt{2}+\sqrt{3}+2\sqrt{2} \\
\end{align}\]
We now write the same type of irrational numbers $2\sqrt{3},\sqrt{3}$ and $-5\sqrt{2},2\sqrt{2}$ close to each other. We have,
\[=2\sqrt{3}+\sqrt{3}-5\sqrt{2}+2\sqrt{2}\]
We take $\sqrt{3},\sqrt{2}$ common and add up their multipliers in bracket to have,
\[\begin{align}
  & =\sqrt{3}\left( 2+1 \right)+\sqrt{2}\left( -5+2 \right) \\
 & =3\sqrt{3}-3\sqrt{2} \\
\end{align}\]
We take 3 common and have the sum as,
\[=3\left( \sqrt{3}-\sqrt{2} \right)\]
(ii) We are given in the question the irrational numbers to add are$\left( 2\sqrt{2}+5\sqrt{3}-7\sqrt{5} \right)$ and $\left( 3\sqrt{3}-\sqrt{2}+\sqrt{5} \right)$. Let us proceed by writing the terms first and opening the bracket following the BODMAS rule.
\[\begin{align}
  & \left( 2\sqrt{2}+5\sqrt{3}-7\sqrt{5} \right)+\left( 3\sqrt{3}-\sqrt{2}+\sqrt{5} \right) \\
 & =2\sqrt{2}+5\sqrt{3}-7\sqrt{5}+3\sqrt{3}-\sqrt{2}+\sqrt{5} \\
\end{align}\]
We now write the terms where the same type of irrational numbers $\sqrt{2},\sqrt{3},\sqrt{5}$ is multiplied close to each other. We have,
\[=2\sqrt{2}-\sqrt{2}+5\sqrt{3}+3\sqrt{3}-7\sqrt{5}+\sqrt{5}\]
We take $\sqrt{2},\sqrt{3},\sqrt{5}$ common and have,
\[\begin{align}
  & =\sqrt{2}\left( 2-1 \right)+\sqrt{3}\left( 5+3 \right)+\sqrt{5}\left( -7+1 \right) \\
 & =\sqrt{2}+8\sqrt{3}-6\sqrt{5} \\
\end{align}\]
The above obtained number is the required sum.\[\]
(iii) We are given in the question the irrational numbers to add are$\left( \dfrac{2}{3}\sqrt{7}-\dfrac{1}{2}\sqrt{2}+6\sqrt{11} \right)$ and $\left( \dfrac{1}{3}\sqrt{7}+\dfrac{3}{2}\sqrt{2}-\sqrt{11} \right)$. Let us proceed by writing both the terms first and open the brackets following BODMAS rule.
\[\begin{align}
  & \left( \dfrac{2}{3}\sqrt{7}-\dfrac{1}{2}\sqrt{2}+6\sqrt{11} \right)+\left( \dfrac{1}{3}\sqrt{7}+\dfrac{3}{2}\sqrt{2}-\sqrt{11} \right) \\
 & =\dfrac{2}{3}\sqrt{7}-\dfrac{1}{2}\sqrt{2}+6\sqrt{11}+\dfrac{1}{3}\sqrt{7}+\dfrac{3}{2}\sqrt{2}-\sqrt{11} \\
\end{align}\]
We now write the terms where the same type of irrational numbers $\sqrt{7},\sqrt{2},\sqrt{11}$ is multiplied close to each other. We have,
\[=\dfrac{2}{3}\sqrt{7}+\dfrac{1}{3}\sqrt{7}-\dfrac{1}{2}\sqrt{2}+\dfrac{3}{2}\sqrt{2}+6\sqrt{11}-\sqrt{11}\]
We take $\sqrt{7},\sqrt{2},\sqrt{11}$ common and have,
\[=\sqrt{7}\left( \dfrac{2}{3}+\dfrac{1}{3} \right)+\sqrt{2}\left( -\dfrac{1}{2}+\dfrac{3}{2} \right)+\sqrt{11}\left( 6-1 \right)\]
We add up the numerators of rational numbers in the brackets as the denominators occurring in them are equal. We have,
\[\begin{align}
  & =\sqrt{7}\left( \dfrac{2}{3}+\dfrac{1}{3} \right)+\sqrt{2}\left( -\dfrac{1}{2}+\dfrac{3}{2} \right)+\sqrt{11}\left( 6-1 \right) \\
 & =\sqrt{7}\left( \dfrac{2+1}{3} \right)+\sqrt{2}\left( \dfrac{-1+3}{2} \right)+\sqrt{11}\left( 5 \right) \\
 & =\sqrt{7}\left( \dfrac{3}{3} \right)+\sqrt{2}\left( \dfrac{2}{2} \right)+5\sqrt{11} \\
 & =\sqrt{7}+\sqrt{2}+5\sqrt{11} \\
\end{align}\]
The above-obtained number is the required sum.\[\]

Note: We note that when we add two irrational numbers most of the time they are irrational but not always. One of the well known example is addition of irrational conjugate numbers $a+b\sqrt{d},a-b\sqrt{d}$ where $a,b,d$ are positive rational numbers and $\sqrt{d}$ is rational. If we add the conjugates we are going to get the sum $2a$, a rational number.