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

2 Answers
Narad T.
Jul 1, 2018

The answer is =3

Explanation:

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

Cem Sentin
Jul 1, 2018

3

Explanation:

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

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