## mathmath333 one year ago question

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{Choose the incorrect relation(s) from the following}\hspace{.33em}\\~\\ & i.)\ \sqrt{6}+\sqrt{2}=\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & ii.)\ \sqrt{6}+\sqrt{2}<\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & iii.)\ \sqrt{6}+\sqrt{2}>\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & a.)\ i.) \ \text{and}\ iii.)\hspace{.33em}\\~\\ & b.)\ ii.) \ \text{and}\ iii.)\hspace{.33em}\\~\\ & c.)\ i.) \hspace{.33em}\\~\\ & d.)\ ii.) \hspace{.33em}\\~\\ \end{align}}

2. mathmath333

i have to solve this with in a minute

3. mathivh

If a calculator is allowed you could just calculate the decimal values of both sides and check if they are equal.

4. mathmath333

lol no calculator is allowed

5. ganeshie8

In view of arriving at a contradiction, lets suppose that $$\sqrt{6}+\sqrt{2}=\sqrt{5}+\sqrt{3}$$. Square both sides and get $$6+2+2\sqrt{12}=5+3+2\sqrt{15}$$. $$\implies \sqrt{12}=\sqrt{15}$$ which is wrong. So the initial assumption is wrong.

6. mathmath333

ok

7. ganeshie8

Alternatively you may work it like this : \large{\begin{align} \sqrt{6}+\sqrt{2}~~&\stackrel{?}{}~~\sqrt{5}+\sqrt{3}\\~\\ (\sqrt{6}+\sqrt{2})^2~~&\stackrel{?}{}~~(\sqrt{5}+\sqrt{3})^2\\~\\ 6+2+2\sqrt{12}~~&\stackrel{?}{}~~ 5+3+2\sqrt{15}\\~\\ \sqrt{12}~~&\stackrel{?}{}~~ \sqrt{15}\\~\\ \sqrt{12}~~&\lt~~ \sqrt{15} \end{align}} since the numbers are positive, the implications will work in the reverse direction also. read it from bottom to top.

8. mathmath333

option 1 is incorrect

9. ganeshie8

i and iii are wrong

10. mathmath333

thnks