Question

$if \: \sqrt{18 - 6 \sqrt{5} } = \sqrt{a} - \sqrt{b} \: then \: prove \: that \: \: \: a + b \: = \: 18$
plzzz help...

Answer 1

Hello Mate!

$\sqrt{18 - 6 \sqrt{5} } = \sqrt{a} - \sqrt{b} \\ on \: squaring \: both \: the \: sides \\ {( \sqrt{18 - 6 \sqrt{5} }) }^{2} = {( \sqrt{a} - \sqrt{b} )}^{2} \\ 18 - 6 \sqrt{5} = a + b - 2 \sqrt{ab} \\ (15 + 3) - 2 \times 3 \times \sqrt{5} = a + b - 2 \sqrt{ab} \\ (15 + 3) - 2 \times \sqrt{3 \times 3 \times 5} = a + b - 2 \sqrt{ab} \\ (15 + 3) - 2 \times \sqrt{45} = a + b - 2 \sqrt{ab} \\ (15 + 3) - 2 \times \sqrt{15} \times \sqrt{3} = a + b - 2 \sqrt{ab} \\ here \: on \: comparing \: we \: get \: \\ 15 + 3 = a + b \\ 18 = a + b \\ hence \: proved$

HOPE IT HELPS☺!