If x > 0 then 1/((2x)^(1/2) + x^(1/2)) = ?

Could you please explain why you multiply and divide by \(\sqrt{2}-1\) ...whats the concept here?

Thanks!

It's called rationalisation and is performed to eliminate irrational expression in the denominator. For this particular case we are doing rationalisation by applying the following rule: \((a-b)(a+b)=a^2-b^2\).

\(\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}\).

Hope it' clear.

Karishma

can you please explain last step . how did you get root x in the numerator

\(\frac{1}{\sqrt{2x} + \sqrt{x}}*\frac{\sqrt{2x} - \sqrt{x}}{\sqrt{2x} - \sqrt{x}} = \frac{\sqrt{2x} - \sqrt{x}}{x}\\
= \frac{\sqrt{x}(\sqrt{2} - 1)}{\sqrt{x}*\sqrt{x}}\) (Take \(\sqrt{x}\) common from the two terms of the numerator. Split the x in the denominator into \(\sqrt{x}*\sqrt{x}\)

Now cancel the \(\sqrt{x}\) from the numerator and denominator.

You get \(\frac{(\sqrt{2} - 1)}{\sqrt{x}}\)

I don't understand this step, could you please explain?

\(\frac{\sqrt{2x}-\sqrt{x}}{x}=\frac{\sqrt{2}\sqrt{x}-\sqrt{x}}{x}=\frac{\sqrt{x}(\sqrt{2}-1)}{x}=\frac{(\sqrt{2}-1)}{\sqrt{x}}\)

What is the last step here? I do not quite understand it

I see there is a lot of confusion about the last step!! It involves rationalizing the denominator. If you multiply and divide by the same number, the expression does not change. So, often when the denominator has irrational numbers/two terms, at least one of which is irrational, we rationalize it by multiplying and dividing by the complementary terms. Say, if the denominator has \(\sqrt{a} - \sqrt{b}\), we multiply and divide the fraction by \(\sqrt{a} + \sqrt{b}\) to get \(\sqrt{a}^2 - \sqrt{b}^2\) (because we know that \((x-y)(x+y) = x^2 - y^2\)) which gives us (a - b) in the denominator (which are rational numbers).

Similarly, here we have

\(\frac{1}{\sqrt{x}(\sqrt{2}+1)}\)

We see that options have \(\sqrt{x}\) in the denominator but no option has \((\sqrt{2}+1)\) in the denominator so we need to get rid of it. So we should multiply and divide by \((\sqrt{2}-1)\) to get

\(\frac{1 * (\sqrt{2}-1)}{\sqrt{x}(\sqrt{2}+1) * (\sqrt{2}-1)} = \\
\frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x} ( 2 - 1)} = \frac{\sqrt{2}-1}{\sqrt{x}}\)

ALTERNATE

Another perspective to solve such question is by substituting values which less people seem to be using but it's very effective

Given: If \(x>0\) then \(\frac{1}{\sqrt{2x}+\sqrt{x}}=?\)

Let's choose a value of x e.g. x=2

then \(\frac{1}{\sqrt{2x}+\sqrt{x}}\)= \(\frac{1}{\sqrt{2*2}+\sqrt{2}}\)
i.e. \(\frac{1}{\sqrt{2x}+\sqrt{x}}\)= \(\frac{1}{2+\sqrt{2}}\)
i.e. \(\frac{1}{\sqrt{2x}+\sqrt{x}}\)= \(\frac{1}{2+1.4}\) = \(\frac{1}{3.4}\) = Approx.\(0.3\)

Check Options @x=2:

a. 1/v(3x) = \(1/\sqrt{6}\) = 1/2.5 INCORRECT [Because the correct value should 1/3.4]
b.1/[2v(2x)] = \(1/2\sqrt{4}\) = 1/4 INCORRECT [Because the correct value should 1/3.4]
c.1/(xv2) = \(1/2\sqrt{2}\) = 1/2.8 INCORRECT [Because the correct value should 1/3.4]
d. (v2-1)/vx = \((1.4-1)/2\sqrt{2}\) = 0.4/1.4 = 1/3.5 CORRECT
e. 1+v2/vx

Answer: option

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