Math, asked by ahirrt, 11 months ago

if a=√8-√7 and a=1/b, than a^2+b^2-3abc=?​

Answers

Answered by samarsparsh18
0

Answer:

a^2=(root8-root7)^2=8+7-2×root(8×7)=15-2root56

a=1/b=>b=1/a=1/(root8-root7)=

1/(root8-root7)×(root8+root7)/(root8+root7)=

root8+root7/(root8)^2-(root7)^2

root8+root7/8-7

root8+root7

b^2=(root8+root7)^2=8+7+2×root(8×7)=15+2root56

Now,

a^2+b^2-3ab=15-2root56+15+2root56+3×a×1/a

=15+15+3=33

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