Math, asked by tanmaytak367, 9 months ago

If x=root3+root2/root3-root 2 &y=1/x,then find the value of x^2+xy-y^2

Answers

Answered by gk730549
5

Step-by-step explanation:

x=√3+√2/√3-√2

rationalise denominator

we have,

x=5+2√6

y=1/5+2√6

again rationalise denominator

y=5-2√6

x^2+xy-y^2

=(5+2√6)^2+(5+2√6)(5-2√6)-(5-2√6)^2

=(49+20√6)+(1)-(49-2√6)

=49+20√6+1-49+20√6

=40√6+1

Answered by unmesh40
1

Answer:

Answer:

Value of x² + y² + xy is 99.

Step-by-step explanation:

Given:

x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\:\:and\:\:y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}x=

3

+

2

3

2

andy=

3

2

3

+

2

To find: value of x² + y² + xy

First we find,

xy=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}xy=

3

+

2

3

2

×

3

2

3

+

2

=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=

(

3

+

2

)(

3

2

)

(

3

2

)(

3

+

2

)

=\frac{(\sqrt{3})^2-(\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}=

(

3

)

2

−(

2

)

2

(

3

)

2

−(

2

)

2

=\frac{3-2}{3-2}=

3−2

3−2

= 1

x^2=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}x

2

=

3

+

2

3

2

×

3

+

2

3

2

=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})^2}=

(

3

+

2

)

2

(

3

2

)

2

=\frac{3+2-2\sqrt{3}\sqrt{2}}{3+2+2\sqrt{3}\sqrt{2}}=

3+2+2

3

2

3+2−2

3

2

=\frac{5-2\sqrt{6}}{5+2\sqrt{6}}=

5+2

6

5−2

6

=\frac{5-2\sqrt{6}}{5+2\sqrt{6}}\times\frac{5-2\sqrt{6}}{5-2\sqrt{6}}=

5+2

6

5−2

6

×

5−2

6

5−2

6

=\frac{(5-2\sqrt{6})^2}{(5+2\sqrt{6})(5-2\sqrt{6})}=

(5+2

6

)(5−2

6

)

(5−2

6

)

2

=\frac{25+24-20\sqrt{6}}{25-24}=

25−24

25+24−20

6

= 49 - 20√6

y^2=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}y

2

=

3

2

3

+

2

×

3

2

3

+

2

=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})^2}=

(

3

2

)

2

(

3

+

2

)

2

=\frac{3+2+2\sqrt{3}\sqrt{2}}{3+2-2\sqrt{3}\sqrt{2}}=

3+2−2

3

2

3+2+2

3

2

=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=

5−2

6

5+2

6

=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}\times\frac{5+2\sqrt{6}}{5+2\sqrt{6}}=

5−2

6

5+2

6

×

5+2

6

5+2

6

=\frac{(5+2\sqrt{6})^2}{(5+2\sqrt{6})(5-2\sqrt{6})}=

(5+2

6

)(5−2

6

)

(5+2

6

)

2

=\frac{25+24+20\sqrt{6}}{25-24}=

25−24

25+24+20

6

= 49 + 20√6

Now,

x² + y² + xy = 49 - 20√6 + 49 + 20√6 + 1 = 49 + 49 + 1 = 99

Therefore, Value of x² + y² + xy is 99

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