Math, asked by sweety13280, 2 months ago

if
x =  \frac{ \sqrt{2 + 1} }{ \sqrt{2 + 1} }
y =  \frac{ \sqrt{2 - 1} }{\sqrt{2 + 1} }
then find the value of x²+y²-34​

Answers

Answered by sujeetgund
0

We have,

x =  \frac{ \sqrt{2 + 1} }{ \sqrt{2 + 1} }  \\  \therefore \: {x}^{2}  =  \frac{2 + 1}{2 + 1}  =  \frac{3}{3}  = 1 \\ Also, \\ y =  \frac{ \sqrt{2 - 1} }{ \sqrt{2 + 1} }  \\  \therefore \:  {y}^{2}  =  \frac{2 - 1}{2 + 1}  =  \frac{1}{3}  \\  \\ So \: now, \:  \\  {x}^{2} +  {y}^{2}  -  34 = 1 +  \frac{1}{3}     -  34 =  \frac{4}{3}   -  34 \\  \\  \implies \: {x}^{2} +  {y}^{2}  -  34 = \frac{ - 98}{3}

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