Math, asked by vbvaish, 1 year ago

if x is equal to root 2 + 1 upon root 2 minus X and Y is equal to root 2 minus 1 upon root 2 + 1 find the value of x squared plus y squared + xy

Answers

Answered by tamannasarawag
182

hope this will help u???

Attachments:
Answered by mysticd
69

Answer:

 Value \:of \: x^{2}+y^{2}+xy = 35

Step-by-step explanation:

 x = \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \:--(1)

and

 y = \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right) \:--(2)

 Value \:of \: x^{2}+y^{2}+xy \\=(x^{2}+y^{2}+2xy)-xy\\=(x+y)^{2}-xy

= \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}+\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)^{2} -\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\times \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)

= \left( \frac{(\sqrt{2}+1)^{2}+(\sqrt{2}-1)^{2}}{(\sqrt{2}-1)(\sqrt{2}+1)}\right)^{2}-1

=\left( \frac{2[(\sqrt{2})^{2}+1^{2}]}{(\sqrt{2})^{2}-1^{2}}\right)^{2}-1

=\left(\frac{2(2+1)}{1}\right)^{2}-1

= (2\times 3)^{2}-1

= 6^{2}-1\\=36-1\\=35

Therefore.,

 Value \:of \: x^{2}+y^{2}+xy = 35

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