# Square root solving

• Feb 16th 2013, 10:46 AM
Amap
Square root solving
If x = 5 + 2√6,
• Feb 16th 2013, 11:28 AM
emakarov
Re: Square root solving
The answer is 2√2. Prove that the square of the expression is 8. To simplify a fraction with the denominator $a+b\sqrt{6}$, multiply the numerator and the denominator by $a-b\sqrt{6}$.
• Feb 16th 2013, 11:36 AM
HallsofIvy
Re: Square root solving
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 $\sqrt{x}$ and do the division directly. However, $\sqrt{x}$ is a little tedious to find so I would look at $\frac{(x- 12)^2}{x}$. $x= 5+ 2\sqrt{6}$ so $x- 1= 4+ 2\sqrt{6}$ and $(x- 1)^2= 16+ 16\sqrt{6}+ 24= 40+ 16\sqrt{6}$. Then $\frac{(x- 1)^2}{x}= \frac{40+ 16\sqrt{6}}{5+ 2\sqrt{6}}$. Do that division by multiplying both numerator and denominator by $5- 2\sqrt{6}$
• Feb 16th 2013, 03:35 PM
Soroban
Re: Square root solving
Hello, Amap!

There is a typo . . .

Quote:

$\text{If }x \,=\, 5 + 2\sqrt{6},\,\text{find }\frac{x-1}{\sqrt{x}}$

$\text{Answer: }\,2{\color{red}\sqrt{2}}$

Note that: . $5 + 2\sqrt{6} \:=\:(\sqrt{3}+\sqrt{2})^2$

Then: . $\frac{x-1}{\sqrt{x}} \;=\;\frac{(5+2\sqrt{6}) -1}{\sqrt{3}+\sqrt{2}} \;=\;\frac{4+2\sqrt{6}}{\sqrt{3} + \sqrt{2}}$

Rationalize: . $\frac{4+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\cdot {\color{blue}\frac{ \sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}} \;=\; \frac{4\sqrt{3} - 4\sqrt{2} + 2\sqrt{18} - 2\sqrt{12}}{3 - 2}$

. . . . . . . . . $=\;4\sqrt{3} - 4\sqrt{2} + 6\sqrt{2} - 4\sqrt{3} \;\;=\;\;\boxed{2\sqrt{2}}$
• Feb 17th 2013, 03:51 AM
ibdutt
Re: Square root solving
Attachment 27081
Quote:

Originally Posted by Amap
If x = 5 + 2√6,