Square root solving

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February 16th 2013, 09:46 AM
Amap

Square root solving

If x = 5 + 2√6,
Find (x-1)/√x ............... Answer is 2√3............. Please explain the steps involved.
February 16th 2013, 10:28 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 $a-b\sqrt{6}$ .
February 16th 2013, 10:36 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 $5- 2\sqrt{6}$
February 16th 2013, 02:35 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: . $\frac{4+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\cdot {\color{blue}\frac{ \sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}} \;=\; \frac{4\sqrt{3} - 4\sqrt{2} + 2\sqrt{18} - 2\sqrt{12}}{3 - 2}$

. . . . . . . . . $=\;4\sqrt{3} - 4\sqrt{2} + 6\sqrt{2} - 4\sqrt{3} \;\;=\;\;\boxed{2\sqrt{2}}$
February 17th 2013, 02:51 AM
ibdutt

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Re: Square root solving

Attachment 27081

Quote:

Originally Posted by Amap

If x = 5 + 2√6,
Find (x-1)/√x ............... Answer is 2√3............. Please explain the steps involved.

The answer is 2 square-root 2