Question

question

Answer 1

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

Answer 2

i have to solve this with in a minute

Answer 3

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

Answer 4

lol no calculator is allowed

Answer 5

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.

Answer 6

ok

Answer 7

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.

Answer 8

option 1 is incorrect

Answer 9

i and iii are wrong

Answer 10

thnks