Question

if x = (root3 + root2)/(root3 - root2) and y = (root3 - root2)/(root3 + root2), find the value of (x + y)^2

Answer 1

$\mathbb{ SOLUTION }$ :

Given,

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \\ \: \: \: \: = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \: \: \: \: \: ( \: by \: \: rationalisation)$

$\: \: \: = \frac{ {( \sqrt{3} + \sqrt{2}) }^{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } \: \: \: ( \therefore \: (a + b)(a - b) = {a}^{2} - {b}^{2} \: \: ) \\ \\ \\ \: \: = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2} \: \: \: \: \: ( \therefore {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} \: ) \: \\ \\ \\ \: \: = \frac{3 + 2 + 2 \sqrt{6} }{1}$
$\: \: = 3 + 2 + 2 \sqrt{6} \\ \\ \\ \therefore \: x \: = 5 + 2 \sqrt{6} \: \: \: \: \: \: \: ........(i)$
Now,

$y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = \frac{1}{x} = \frac{1}{5 + 2 \sqrt{6} } \: \: \: (from \: \: eq. \: (i) \: ) \\ \\ \\ \: \: = \frac{1}{5 + 2 \sqrt{6} } \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } \: \: \: \: \: (by \: \: rationalisation) \\ \\ \\ \: \: = \frac{5 - 2 \sqrt{6} }{ {(5)}^{2} - {(2 \sqrt{6}) }^{2} } \: \: \: ( \therefore \: (a - b)(a + b) = {a}^{2} - {b}^{2} \: ) \\ \\ \\ \: \: = \frac{5 - 2 \sqrt{6} }{25 - 24} \\ \\ \\ y = \frac{5 - 2 \sqrt{6} }{1} = 5 - 2 \sqrt{6}$


Now,

$x + y = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} = 10 \\ \\ \\ on \: \: squaring \: \: both \: \: sides \: \\ we \: \: get \\ \\ {(x + y)}^{2} = {(10)}^{2} = 100$

Answer 2

$<b>Heya mate. (^_-). Solution below.$
====================================

♦ x = (√3 + √2)/(√3 - √2)


→ Rationalising the denominator,


=> x = [(√3 + √2)/(√3 - √2)] × [(√3 + √2)/(√3 + √2)] ... [multiplying by the conjugate]


=> x = [(√3 + √2)(√3 + √2)]/[(√3 - √2)/(√3 + √2)]


• Using identity : (a - b)(a + b) = a² - b² in the denominator,


=> x = (√3 + √2)²/(√3)² - (√2)²


• Using identity : (a + b)² = a² + b² + 2ab in the numerator,


=> x = (3 + 2 + 2 × √3 × √2)/(3 - 2)

=> x = 5 + 2√6/1

=> x = 5 + 2√6


•°• x = 5 + 2√6 ... [equation i]




♦ y = (√3 - √2)/(√3 + √2)


→ Rationalising the denominator,


=> y = [(√3 - √2)/(√3 + √2)] × [(√3 - √2)/(√3 - √2)]

=> y = [(√3 - √2)(√3 - √2)]/[(√3 + √2)(√3 - √2)]


• Using identity : (a + b)(a - b) = a² - b² in the denominator,


=> y = (√3 - √2)²/(√3)² - (√2)²


• Using identity : (a - b)² = a² + b² - 2ab in the numerator,


=> y = (3 + 2 - 2 × √3 × √2)/3 - 2

=> y = 5 - 2√6/1

=> y = 5 - 2√6


•°• y = 5 - 2√6 ... [equation ii]



♦ Adding equations i and ii,

x + y = (5 + 2√6) + (5 - 2√6)

=> x + y = 5 + 2√6 + 5 - 2√6

=> x + y = 10 ... (2√6 and -2√6 get cancelled)


•°• x + y = 10... [equation iii]



♦ Now, squaring both sides in equation iii,

x + y = 10

=> (x + y)² = (10)²

=> (x + y)² = 100


•°• (x + y)² = 100
====================================

$<marquee>Thank you.</marquee>$