Question

find x where x is positive number​

Answer 1

Answer:

$2\sqrt{5}$

Step-by-step explanation:

$x=0$ is not a root, so both sides of $x^2+4x-1=0$ can be divided by $x$.

• $x-\frac{1}{x} =-4$

• $x*\frac{1}{x} =1$

Main Key : $(x+\frac{1}{x} )^2=(x-\frac{1}{x} )^2+4x*\frac{1}{x}$

$(x+\frac{1}{x} )^2=16+4=20$

It is given that $x>0$.

Then, $x+\frac{1}{x}>0$.

∴$x+\frac{1}{x} =2\sqrt{5}$