Radical Equation Solution: Simplifying and Transposing for Accurate Results?

[paulmdrdo1]

$\sqrt{x}-\frac{2}{\sqrt{x}}=1$

transposing $\frac{2}{\sqrt{x}}$ to the right-side

$\sqrt{x}=\frac{2}{\sqrt{x}}$ multiply by $\sqrt{x}$

I get,

$x=2$ is my solution correct? if not please explain why. thanks!

 

[kaliprasad]

$\sqrt{x}-\frac{2}{\sqrt{x}}=1$

transposing $\frac{2}{\sqrt{x}}$ to the right-side

$\sqrt{x}=\frac{2}{\sqrt{x}}$ multiply by $\sqrt{x}$

I get,

$x=2$ is my solution correct? if not please explain why. thanks!

It is not right. because you dropped the 1

you will get
$\sqrt{x}=1+ \frac{2}{\sqrt{x}}$

but this approach shall not help

1st step is to multiply by $\sqrt{x}$ to remove redical from denominator

 

[Maged Saeed]

$\sqrt{x}-\frac{2}{\sqrt{x}}=1$

transposing $\frac{2}{\sqrt{x}}$ to the right-side

$\sqrt{x}=\frac{2}{\sqrt{x}}$ multiply by $\sqrt{x}$

I get,

$x=2$ is my solution correct? if not please explain why. thanks!

you can solve it in that way :
$$\sqrt{x}-\frac{2}{\sqrt{x}}=1$$
multiply by $$\sqrt{x}$$ and moving the other term to the other side :
$$x=1+2=3$$
hope that is right ..
=)

 

[Deveno]

you can solve it in that way :
$$\sqrt{x}-\frac{2}{\sqrt{x}}=1$$
multiply by $$\sqrt{x}$$ and moving the other term to the other side :
$$x=1+2=3$$
hope that is right ..
=)

No, that isn't right, either.

If we multiply the equation:

$\sqrt{x} - \dfrac{2}{\sqrt{x}} = 1$

by $\sqrt{x}$, we obtain:

$x - 2 = \sqrt{x}$

Now, the best thing to do is square both sides. This will introduce an "extraneous" solution, so be sure to test both answers against the original problem.

 

[alyafey22]

I suggest letting $x=t^2$ if you don't prefer working with radicals , just like me.