Question

If x = √2 + 1/√2 - 1 and y = √2 - 1/ √2+1 find value of x^2 y^2 + xy

Answer 1

Answer:

2

Explanation:

x=(√2+1)/(√2-1)

y= (√2-1)/(√2+1)

Now,

xy = [(√2+1)/(√2-1)][(√2-1)/(√2+1)]

=> xy = 1

Therefore,

x²y²+xy

= (xy)²+xy

= 1²+1

= 2

••••

Answer 2

$\huge\underline\mathfrak{Answer:}$

We have

$x = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1} \\ \\ = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1 } \\ \\ = \frac{( \sqrt{2} + 1) {}^{2} }{( \sqrt{2}) {}^{2} - (1) {}^{2} } \\ \\ = \frac{2 + 1 + 2 \sqrt{2} }{2 - 1} \\ \\ = 3 + 2 \sqrt{2} \\ \\ y = \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } \\ \\ = \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1} \\ \\ = \frac{( \sqrt{2} - 1) {}^{2} }{( \sqrt{2} ) {}^{2} - {(1)}^{2} } \\ \\ = \frac{2 + 1 - 2 \sqrt{2} }{2 - 1} \\ \\ = 3 - 2 \sqrt{2}$

•°• x + y = ( 3 + 2√2) + (3-2√2)

$\implies$ x + y = 6

and xy = ( 3 + 2√2)(3-2√2)

$\implies$xy = (3)² - (2√2)²

$\implies$ xy = 9 - 8 = 1

Now, x² + y² + xy = (x² + y² + 2xy) - xy

= (x + y)² - xy

= (6)² - (1)

= 36 - 1

= 35.