Question

if x = √3+√2/√3-√2 and y = √3-√2/√3+√2 then find the values (i) x² + y² (ii) x³ + y³ please answer both with step by steps.​

Answer 1

Step-by-step explanation:

hope this helps you better

Answer 2

Given :

• $\sf\:x~=~\frac{\sqrt{3} +\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

• $\sf\:y~=~\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}+\sqrt{2}}$

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To find :

Value of -

• x² + y²

• x³ + y³

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Concept :

Here in the question we are given the value of x and y. So calculating the value of ( x² + y² ) and ( x³ + y³ ) is simple, we just need to substitute the given values of x and y in the expressions. But we would first simplify the values of x and y by Rationalizing them.

In order to rationalize we need to multiply the given ratio by the conjugate of the denominator. So after rationalizing and substituting the rationalized value of x and y in the expression we would get our required answer.

Hope am clear let's solve :D~

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Solution :

Rationalizing the value of x -

$\sf\:x~=~\frac{\sqrt{3} +\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

We know conjugate of ( √3-√2 ) is ( √3+√2 ) so multiplying the ratio by ( √3+√2 )

$\sf\implies\:x~=~\frac{\sqrt{3} +\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3} +\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$\sf\implies\:x~=~\frac{(\sqrt{3} +\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

$\sf\implies\:x~=~\frac{(\sqrt{3}) ^{2} +2 \times \sqrt{3} \times \sqrt{2} +(\sqrt{2})^{2}}{3-2}$

$\sf\implies\:x~=~\frac{3+2\sqrt{6} +2}{1}$

$\sf\implies\:x~=~\frac{3+2+2\sqrt{6}}{1}$

$\sf\color{teal}{\implies\:x~=~5+2\sqrt{6}}$

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Rationalizing the value of y -

$\sf\:y~=~\frac{\sqrt{3} -\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

We know conjugate of ( √3+√2 ) is ( √3-√2 ) so multiplying the ratio by ( √3-√2 )

$\sf\implies\:y~=~\frac{\sqrt{3} -\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3} -\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$\sf\implies\:y~=~\frac{(\sqrt{3} -\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$

$\sf\implies\:y~=~\frac{(\sqrt{3}) ^{2} -2 \times \sqrt{3} \times \sqrt{2} +(\sqrt{2})^{2}}{3-2}$

$\sf\implies\:y~=~\frac{3-2\sqrt{6} +2}{1}$

$\sf\implies\:y~=~\frac{3+2-2\sqrt{6}}{1}$

$\sf\color{teal}{\implies\:y~=~5-2\sqrt{6}}$

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Calculating the required value for the expressions -

(i) $\sf\:x^{2}+y^{2}$

Substituting the simplified values of x and y

$\sf\implies\:(5+2\sqrt{6})^{2}+(5-2\sqrt{6})^{2}$

Using an identities :

• ( a + b )² = a² + 2ab + b²

• ( a - b )² = a² - 2ab + b²

$\sf\implies\:[(5)^{2} +2 \times 5 \times 2\sqrt{6}+ (2\sqrt{6})^{2}]+(5-2\sqrt{6})^{2}$

$\sf\implies\:[25 +20\sqrt{6} + 4 \times 6]+[(5)^{2} - 2 \times 5 \times 2\sqrt{6} +(2\sqrt{6})^{2}$

$\sf\implies\:[25 +20\sqrt{6} + 4 \times 6]+[25 -20\sqrt{6} + 4 \times 6]$

$\sf\implies\:[25 +20\sqrt{6}+ 24]+[25 -20\sqrt{6}+24]$

Multiplying the signs and removing the brackets

$\sf\implies\:25 +20\sqrt{6}+ 24+25 -20\sqrt{6}+24$

Arranging the numbers

$\sf\implies\:25 +24+25+24+20\sqrt{6} -20\sqrt{6}$

Proceeding simple calculation

$\sf\implies\:49+25+24+20\sqrt{6} -20\sqrt{6}$

$\sf\implies\:74+24+20\sqrt{6} -20\sqrt{6}$

$\sf\implies\:98+20\sqrt{6} -20\sqrt{6}$

$\sf\implies\:98+\cancel{20\sqrt{6}}-\cancel{20\sqrt{6}}$

$\small{\underline{\boxed{\mathrm{\implies\:98}}}}$ $\tt\color{blue}{\bigstar}$

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(ii) $\sf\:x^{3}+y^{3}$

Using an identity :

• a³ + b³ = ( a + b ) ( a² - ab + b² )

$\sf\implies\:(x+y)(x^{2}-xy+y^{2})$

Substituting the values

$\sf\implies\:[5+2\sqrt{6}+(5-2\sqrt{6})](x^{2}-xy+y^{2})$

$\sf\implies\:[5+2\sqrt{6}+(5-2\sqrt{6})][(x^{2}+y^{2})-xy]$

$\sf\implies\:[5+2\sqrt{6}+5-2\sqrt{6}][98-xy]$

$\sf\implies\:[10+2\sqrt{6}-2\sqrt{6}][98-xy]$

$\sf\implies\:[10+\cancel{2\sqrt{6}}-\cancel{2\sqrt{6}}][98-xy]$

$\sf\implies\:10[98-xy]$

$\sf\implies\:10[98-(5+2\sqrt{6}) (5-2\sqrt{6})]$

$\sf\implies\:10[98-(5)^{2}-(2\sqrt{6})^{2}]$

$\sf\implies\:10[98-25+4\times 6]$

$\sf\implies\:10[98-25+24]$

$\sf\implies\:10[98-1]$

$\sf\implies\:10[97]$

$\sf\implies\:10 \times 97$

$\small{\underline{\boxed{\mathrm{\implies\:970}}}}$ $\tt\color{blue}{\bigstar}$

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Henceforth -

❒ x² + y² = $\large{\mathfrak\purple{98}}$

❒ x³ + y³ = $\large{\mathfrak\purple{970}}$

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