If #x^2+1/x^2=47#, then #sqrtx+1/sqrtx=#?

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Answer 1 · 2018-07-01T14:39:22

The answer is #=3#

If

#x^2+1/x^2=47#

#(x+1/x)^2=x^2+1/x^2+2*x*1/x=x^2+1/x^2+2#

Substituting the value of #x^2+1/x^2=47#

#(x+1/x)^2=47+2=49#

#x+1/x=sqrt(49)=7#

#(sqrtx+1/sqrtx)^2=(sqrtx)^2+(1/sqrtx)^2+2*sqrtx*1/sqrtx#

#=x+1/x+2#

#=7+2=9#

Therefore,

#(sqrtx+1/sqrtx)^2=9#

#=>#, #sqrtx+1/sqrtx=sqrt9=3#

Answer 2 · 2018-07-01T18:43:20

#3#

Set #y=sqrtx+1/sqrtx#

Hence,

#y^4=(sqrtx+1/sqrtx)^4#

#y^4=x^2+4x+6+4/x+1/x^2#

#y^4=x^2+1/x^2+4*(x+1/x)+6#

#y^4=47+4*((sqrtx+1/sqrtx)^2-2)+6#

#y^4=4*(y^2-2)+53#

#y^4=4y^2+45#

#y^4-4y^2-45=0#

#(y^2+5)*(y^2-9)=0#

Thus, #y^2=9#, so #y=sqrtx+1/sqrtx=3#