# Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$

If $n=7+4\sqrt3$,then what is the simplified value of $$\sqrt n+\frac {1}{\sqrt n}$$ I was taking LCM but how to get rid of $\sqrt n$ in denominator

Hint: We have $\left(\sqrt n+\dfrac {1}{\sqrt n}\right)^2=n+2+\frac 1n$, then multiply top and bottom of the last by the conjugate.

Square $2 + \sqrt{3}$, and you will know what to do.

Ross Millikan's answer is much more elegant, but I am posting this answer as well because it is in a sense more general. (By this I mean it will work for any expression involving $\sqrt{n}$, and not just this epxression in particular.) First, we make a guess that $$\sqrt{n} = a + b\sqrt{3}$$

Squaring both sides, we find that $$7 + 4\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3}$$

Can you think of $a$ and $b$ integers that will make $a^2 + 3b^2 = 7$ and $2ab = 4$? If so, you can write $\sqrt{n}$ in a nicer form and the expression should be simple to calculate.

(Note that this method doesn't always work. But in general if you ever have a square root of a number of the form $x + y\sqrt{z}$, it can't hurt to try making $\sqrt{x + y\sqrt{z}} = x' + y'\sqrt{z}$.)

• This is new one.Is there any other replacement like this for similar type of expression May 17, 2013 at 0:11

You could as OP suggests, add 2, and take the square root.

For example, $\dfrac{1}{(7+4\sqrt3)}$ is $(7-4 \sqrt3)$, so the sum of these numbers is $14$. You add $2$ to it to get $16$, and take the square root.

The new number is $4 = x + \dfrac {1}x$, which leads to $x = \dfrac{\sqrt{4+2} + \sqrt{4-2}}{2}$ = 1.93185165259.

• The new number is $4 = x + 1/x$, not (x+1)/x. May 17, 2013 at 8:14