Question

if p =root3 - root 2 /root3 + root 2 and q =root 3 +root 2 /root 3 -root 2 find p square +q square

Answer 1

Hi...☺

Here is your answer...✌

GIVEN THAT

$p = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3 } + \sqrt{2} } \: \: and\: \: q = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$
We know that,

(p + q)² = p² + q² + 2pq

=> (p + q)² - 2pq = p² + q²

=> p² + q² = (p + q)² - 2pq .....(1)

First we calculate

${(p + q)}^{2} = {( \frac{ \sqrt{3 } - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } + \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } )}^{2} \\ \\ = {(\frac{{ (\sqrt{3} - \sqrt{2})}^{2} + {( \sqrt{3} + \sqrt{2} )}^{2} }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} ) } )}^{2}\\ \\ =({ \frac{3 + 2 - 2 \sqrt{6} + 3 + 2 + 2 \sqrt{6} }{ {(\sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} })}^{2} \\ \\ = {(\frac{10}{3 - 2}) }^{2}= {(\frac{10}{1} )}^{2}= {10}^{2}=100$

Now,

We calculate

$p \times q = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ pq = 1$

Putting value of (p+q)² and pq in (1)
We get,

p² + q² = 100-2×1 = 100-2 = 98

=> p² + q² = 98

Answer 2

Hey there!

Solution is attached !

The answer is 98