Question

Find the square root of \[n+\sqrt{{{n}^{2}}-1}\]

Answer 1

Hint: In this problem we have to find the square root of \[n+\sqrt{{{n}^{2}}-1}\] for that assign with \[{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}\]
And use the property we will get the equation in the form of \[n+\sqrt{{{n}^{2}}-1}\]
And compare the values. Solve and simplify it to get the square root of \[n+\sqrt{{{n}^{2}}-1}\] .

Complete step-by-step answer:
In this particular problem we have to find the square root of \[n+\sqrt{{{n}^{2}}-1}\]
Square root of \[n+\sqrt{{{n}^{2}}-1}\] is in the form of \[\sqrt{a}+\sqrt{b}\]
Therefore, \[n+\sqrt{{{n}^{2}}-1}={{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}--(1)\] .
By squaring this equation \[\sqrt{a}+\sqrt{b}\] we get in the form of \[n+\sqrt{{{n}^{2}}-1}\] so, we compare the values then we get:
 \[{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}\]
Apply the property of \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] in this above equation
 \[{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}={{\left( \sqrt{a} \right)}^{2}}+2\sqrt{a}\sqrt{b}+{{\left( \sqrt{b} \right)}^{2}}\]
By simplifying further we get:
 \[{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}=a+2\sqrt{ab}+b\]
Arrange the term so, that we can compare with \[n+\sqrt{{{n}^{2}}-1}\]
 \[{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}=a+b+2\sqrt{ab}\]
Substitute this value in equation (1) we get:
 \[n+\sqrt{{{n}^{2}}-1}=a+b+2\sqrt{ab}\]
Compare this and you will get two equations.
 \[a+b=n---(2)\]
 \[2\sqrt{ab}=\sqrt{{{n}^{2}}-1}---(3)\]
From the equation (1) we will get the value of b
 \[b=n-a\]
In equation (2) square on both sides we get:
 \[{{\left( 2\sqrt{ab} \right)}^{2}}={{\left( \sqrt{{{n}^{2}}-1} \right)}^{2}}\]
By simplifying we get:
 \[4ab={{n}^{2}}-1\]
Substitute the value of b in above equation then we get:
 \[4a(n-a)={{n}^{2}}-1\]
After simplifying and further solving this we get:
 \[4an-4{{a}^{2}}-{{n}^{2}}+1=0\]
We have to arrange the term and write in the quadratic form
 \[4{{a}^{2}}-4an+\left( {{n}^{2}}-1 \right)=0\]
For general formula for roots of quadratic form is
 \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
But in this problem variable is a
Therefore, \[a=\dfrac{-(-4n)\pm \sqrt{{{(-4n)}^{2}}-4\left( 4 \right)\left( {{n}^{2}}-1 \right)}}{2\times 4}\]
After simplifying and further solving we get:
 \[a=\dfrac{4n\pm \sqrt{16{{n}^{2}}-4\left( 4 \right)\left( {{n}^{2}}-1 \right)}}{8}\]
Further solving we get:
 \[a=\dfrac{4n\pm 4\sqrt{{{n}^{2}}-\left( {{n}^{2}}-1 \right)}}{8}\]
Here, 4 gets common and cancelled with denominator
 \[a=\dfrac{n\pm \sqrt{{{n}^{2}}-\left( {{n}^{2}}-1 \right)}}{2}\]
Further simplifying by opening the brackets
 \[a=\dfrac{n\pm \sqrt{{{n}^{2}}-{{n}^{2}}+1}}{2}\]
And solve this further to get the value of a
 \[a=\dfrac{n\pm 1}{2}\]
So we get two values of a
That is \[a=\dfrac{n+1}{2}\] and \[a=\dfrac{n-1}{2}\]
For \[a=\dfrac{n+1}{2}\] substitute this in equation (2) to get the value of b
 \[b=n-\dfrac{n+1}{2}\]
After simplifying we get:
 \[b=\dfrac{n-1}{2}\]
Now, square root of \[n+\sqrt{{{n}^{2}}-1}\] is \[\sqrt{a}+\sqrt{b}\]
Substitute the value of a and b in this above equation
 \[\sqrt{n+\sqrt{{{n}^{2}}-1}}=\sqrt{\dfrac{n+1}{2}}+\sqrt{\dfrac{n-1}{2}}\]
For \[a=\dfrac{n-1}{2}\] substitute this in equation (2) to get the value of b
 \[b=n-\dfrac{n-1}{2}\]
After simplifying we get:
 \[b=\dfrac{n+1}{2}\]
Now, square root of \[n+\sqrt{{{n}^{2}}-1}\] is \[\sqrt{a}+\sqrt{b}\]
Substitute the value of a and b in this above equation
 \[\sqrt{n+\sqrt{{{n}^{2}}-1}}=\sqrt{\dfrac{n-1}{2}}+\sqrt{\dfrac{n+1}{2}}\]
So, there are two square root of \[n+\sqrt{{{n}^{2}}-1}\] which can also be written as \[\sqrt{n+\sqrt{{{n}^{2}}-1}}=\sqrt{\dfrac{n+1}{2}}+\sqrt{\dfrac{n-1}{2}}\]
As well as \[\sqrt{n+\sqrt{{{n}^{2}}-1}}=\sqrt{\dfrac{n-1}{2}}+\sqrt{\dfrac{n+1}{2}}\]

Note: In this particular problem keep in mind while finding the \[\sqrt{n+\sqrt{{{n}^{2}}-1}}\] it is always in the form of \[\sqrt{a}+\sqrt{b}\] . Don’t make silly mistake while simplifying and remember that I f we get two values of a then we need to find the two square roots of \[n+\sqrt{{{n}^{2}}-1}\] you can see in above solution. So, the above solution can be preferred for such types of problems.