1.

✓√2-1/✓2+1 answer my question as much faster you can​

Answer»

We have to find the value of  \sqrt{\dfrac{\sqrt2-1}{\sqrt2+1}}.

First, let's rationalize the denominator.

For this, we can multiply  \sqrt2-1  to both sides.

\sqrt{\dfrac{\sqrt2-1}{\sqrt2+1}}\ \Longrightarrow\ \sqrt{\dfrac{(\sqrt2-1)(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}}

Here the numerator becomes a square and the denominator gets simplified according to  (a-b)(a+b)=a^2-b^2.

\sqrt{\dfrac{(\sqrt2-1)(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}}\ \Longrightarrow\ \sqrt{\dfrac{(\sqrt2-1)^2}{(\sqrt2)^2-(1)^2}}\ \Longrightarrow\ \sqrt{\dfrac{(\sqrt2-1)^2}{2-1}}\ \Longrightarrow\ \\ \\ \\ \\ \Longrightarrow\ \sqrt{\dfrac{(\sqrt2-1)^2}{1}}\ \Longrightarrow\ \sqrt{(\sqrt2-1)^2}

Now we can take the square root of it. But before, we should have a think.

On taking the square root of a number, we can GET two corresponding values with sign difference, as an example, the square root of 4 is 2 as well as -2.

But if the square root of a number has to be found by indicating that number under the SYMBOL "√",  we can only take the "positive" square root of that number.

Because the symbol "√" means "positive square root", not "square root" only.

To get both square roots, both positive and negative, we may INDICATE the number under "±√".

Here it is \sqrt{(\sqrt2-1)^2},  so we have to get the positive square root.

On FINDING \pm\sqrt{(\sqrt2-1)^2},  we get  \sqrt2-1  and  -(\sqrt2-1)=1-\sqrt2.

Now we have to find which is positive among these.

For this, we consider  2 > 1.

2>1\ \Longrightarrow\ \sqrt2\ >\sqrt1\ \Longrightarrow\ \sqrt2\ >1\ \Longrightarrow\ \sqrt2-1>0

So  \sqrt2-1  is positive, hence  \sqrt2-1  is the answer.



Discussion

No Comment Found

Related InterviewSolutions