Question

If x=√5+√2÷√5-√2 and y=√5-√2÷√5+√2 find 3x^2+4xy-3y^2

Answer 1

here is your answer...
hope this helps you...

Answer 2

Answer:

Required value is $\frac{ - 56 \sqrt{10 } + 12}{3}$

Step-by-step explanation:

Given,

$x = \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

and

$y = \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} }$

We want to find,

$3 {x}^{2} + 4xy - 3 {y}^{2}$

$3( {x}^{2} - {y}^{2} ) + 4xy$

$3(x + y)(x - y) + 4xy$

Now,

${x}^{2} = \frac{ { (\sqrt{5} - \sqrt{2}) }^{2} }{ { (\sqrt{5} + \sqrt{2} )}^{2} } = \frac{7 - 2 \sqrt{10} }{7 + 2 \sqrt{10} }$

${y}^{2} = \frac{ { (\sqrt{5} + \sqrt{2}) }^{2} }{ { (\sqrt{5} - \sqrt{2} )}^{2} } = \frac{7 + 2 \sqrt{10} }{7 - 2 \sqrt{10} }$

So,

${x}^{2} - {y}^{2} = \frac{7 - 2 \sqrt{10} }{7 + 2 \sqrt{10} } - \frac{7 + 2 \sqrt{10} }{7 - 2 \sqrt{10} } = \frac{49 - 28 \sqrt{10} + 40 - 49 - 28 \sqrt{10 }- 40}{49 - 40} = - \frac{56 \sqrt{10} }{9}$

And

$xy = \frac{ \sqrt{5} + \sqrt{2} }{ \sqrt{5} - \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} + \sqrt{2} } = 1$

Now,

$3( {x}^{2} - {y}^{2} ) + 4xy = 3( - \frac{56 \sqrt{10} }{9} ) + 4 = - \frac{56 \sqrt{10} }{3} + 4 = \frac{ - 56 \sqrt{10 } + 12}{3}$