If x = 5 + 2√6,
Find (x-1)/√x ............... Answer is 2√3............. Please explain the steps involved.
First, no, that is not the answer. Are you sure you copied it correctly?
It would be possible, since you are given x, to find both x- 1 and $\displaystyle \sqrt{x}$ and do the division directly. However, $\displaystyle \sqrt{x}$ is a little tedious to find so I would look at $\displaystyle \frac{(x- 12)^2}{x}$. $\displaystyle x= 5+ 2\sqrt{6}$ so $\displaystyle x- 1= 4+ 2\sqrt{6}$ and $\displaystyle (x- 1)^2= 16+ 16\sqrt{6}+ 24= 40+ 16\sqrt{6}$. Then $\displaystyle \frac{(x- 1)^2}{x}= \frac{40+ 16\sqrt{6}}{5+ 2\sqrt{6}}$. Do that division by multiplying both numerator and denominator by $\displaystyle 5- 2\sqrt{6}$
Hello, Amap!
There is a typo . . .
$\displaystyle \text{If }x \,=\, 5 + 2\sqrt{6},\,\text{find }\frac{x-1}{\sqrt{x}}$
$\displaystyle \text{Answer: }\,2{\color{red}\sqrt{2}} $
Note that: .$\displaystyle 5 + 2\sqrt{6} \:=\:(\sqrt{3}+\sqrt{2})^2$
Then: .$\displaystyle \frac{x-1}{\sqrt{x}} \;=\;\frac{(5+2\sqrt{6}) -1}{\sqrt{3}+\sqrt{2}} \;=\;\frac{4+2\sqrt{6}}{\sqrt{3} + \sqrt{2}} $
Rationalize: .$\displaystyle \frac{4+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\cdot {\color{blue}\frac{ \sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}} \;=\; \frac{4\sqrt{3} - 4\sqrt{2} + 2\sqrt{18} - 2\sqrt{12}}{3 - 2}$
. . . . . . . . . $\displaystyle =\;4\sqrt{3} - 4\sqrt{2} + 6\sqrt{2} - 4\sqrt{3} \;\;=\;\;\boxed{2\sqrt{2}}$