State both Values of x that satisfy the equation....

Solve for x:
7-2/x = sqrt(7) sqrt(1/x)

Multiply both sides by x:
7 x-2 = sqrt(7)/sqrt(1/x)

Multiply both sides by sqrt(1/x):
7/sqrt(1/x)-2 sqrt(1/x) = sqrt(7)

Square both sides:
(7/sqrt(1/x)-2 sqrt(1/x))^2 = 7

(7/sqrt(1/x)-2 sqrt(1/x))^2 = 49 x-28+4/x:
49 x-28+4/x = 7

Bring 49 x-28+4/x together using the common denominator x:
(49 x^2-28 x+4)/x = 7

Multiply both sides by x:
49 x^2-28 x+4 = 7 x

Subtract 7 x from both sides:
49 x^2-35 x+4 = 0

The left hand side factors into a product with two terms:
(7 x-4) (7 x-1) = 0

Split into two equations:
7 x-4 = 0 or 7 x-1 = 0

Add 4 to both sides:
7 x = 4 or 7 x-1 = 0

Divide both sides by 7:
x = 4/7 or 7 x-1 = 0

Add 1 to both sides:
x = 4/7 or 7 x = 1

Divide both sides by 7:
x = 4/7 or x = 1/7

7-2/x ⇒ 7-2/(1/7) = -7
sqrt(7) sqrt(1/x) ⇒ sqrt(7) sqrt(1/(1/7)) = 7:
So this solution is incorrect

7-2/x ⇒ 7-2/(4/7) = 7/2
sqrt(7) sqrt(1/x) ⇒ sqrt(7) sqrt(1/(4/7)) = 7/2:
So this solution is correct

The solution is:
Answer: | x = 4/7

(7-2/x)=sqrt(7/x)

 

\(7-\frac{2}{x}=\sqrt{\frac{7}{x}}\\ (7-\frac{2}{x})^2=\frac{7}{x}\\ 49-\frac{28}{x}+(\frac{2}{x})^2=\frac{7}{x}\\ 49+\frac{4}{x^2}=\frac{35}{x}\\ 49x^2+4=35x\\ 49x^2-35x+4=0\\ x=\frac{35\pm\sqrt{1225-16*49}}{98}\\ x=\frac{35\pm21}{98}\\ x=\frac{56}{98}\qquad or \qquad x=\frac{14}{98}\\ x=\frac{4}{7}\qquad or \qquad x=\frac{1}{7}\\ \)

 

but  

\(x>0\;\;and \;\;\\ 7-\frac{2}{x}>0\\ 7>\frac{2}{x}\\ 7x>2\\ x>2/7\\~\\ \therefore x=\frac{4}{7}\)