Question

if 3 root 2 +2 root 3 \5root2-4root 3 =x-y root 6then find the value of x and y

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{3 \sqrt{2} + 2 \sqrt{3} }{5 \sqrt{2} - 4 \sqrt{3} } = x - y \sqrt{6}$

So, on rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{3 \sqrt{2} + 2 \sqrt{3} }{5 \sqrt{2} - 4 \sqrt{3} } \times \dfrac{5 \sqrt{2} + 4 \sqrt{3} }{5 \sqrt{2} + 4 \sqrt{3} } = x - y \sqrt{6}$

We know,

$\boxed{ \tt{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{3 \sqrt{2}(5 \sqrt{2}+4 \sqrt{3}) + 2 \sqrt{3}(5 \sqrt{2} + 4 \sqrt{3})}{ {(5 \sqrt{2} )}^{2} - {(4 \sqrt{3})}^{2} } = x - y \sqrt{6}$

$\rm :\longmapsto\:\dfrac{30 + 12 \sqrt{6} + 10 \sqrt{6} + 24}{50 - 48} = x - y \sqrt{6}$

$\rm :\longmapsto\:\dfrac{54 + 22 \sqrt{6}}{2} = x - y \sqrt{6}$

$\rm :\longmapsto\:\dfrac{2(27 + 11 \sqrt{6})}{2} = x - y \sqrt{6}$

$\rm \implies\: \purple{27} + \pink{11 \sqrt{6}} \: = \: \purple{x} - \pink{y \sqrt{6}}$

On Comparing, we get

$\rm :\longmapsto\:\boxed{ \tt{ \: x = 27 \: \: and \: y \: = \: - \: 11 \: }}$

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Explore More :-

↝ More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)