Surds (Q3)

**Punch** · August 20th 2010, 06:58 AM

Without using a calculator, find the values of the integers $a$ and $b$ for which the solution of the equation

$x\sqrt{24} = x\sqrt{3} + \sqrt{6}$

is $\frac{a+\sqrt{b}}{7}$

**mathaddict** · August 20th 2010, 07:37 AM

Originally Posted by Punch

Without using a calculator, find the values of the integers $a$ and $b$ for which the solution of the equation

$x\sqrt{24} = x\sqrt{3} + \sqrt{6}$

is $\frac{a+\sqrt{b}}{7}$

$x=\frac{\sqrt{6}}{\sqrt{24}\sqrt{3}}$

Rationalize the surds and you will get

$\frac{\sqrt{144}+\sqrt{18}}{21}$

$\frac{12+3\sqrt{2}}{21}<br />$
...

**Soroban** · August 20th 2010, 10:50 AM

Hello, Punch!

$\text{Without using a calculator, find the values of }a\text{ and }b$

$\text{for which the solution of: }\:x\sqrt{24} \:=\: x\sqrt{3} + \sqrt{6}$

. . $\text{is: }\;x\:=\:\dfrac{a+\sqrt{b}}{7}$

We have: . $x\sqrt{24} \;=\;x\sqrt{3} + \sqrt{6} \quad\Rightarrow\quad x\sqrt{24} - x\sqrt{3} \;=\;\sqrt{6}$

. . . . . $(\sqrt{24}-\sqrt{3})x \;=\;\sqrt{6} \quad\Rightarrow\quad x \;=\;\dfrac{6}{\sqrt{24}-\sqrt{3}}$

Factor and reduce:

. . $x \;=\;\dfrac{\sqrt{2}\!\cdot\!\sqrt{3}}{2\!\cdot\!\ sqrt{2}\!\cdot\!\sqrt{3} - \sqrt{3}} \;=\;\dfrac{\sqrt{3}\!\cdot\!\sqrt{2}} {\sqrt{3}\!\cdot\!(2\sqrt{2}-1)} \;=\; \dfrac{\sqrt{2}}{2\sqrt{2}-1}$

Rationalize:

. . $x \;=\;\dfrac{\sqrt{2}}{2\sqrt{2}-1}\cdot\dfrac{2\sqrt{2}+1}{2\sqrt{2}+1} \;=\;\dfrac{4 + \sqrt{2}}{8-1} \;=\;\dfrac{4 + \sqrt{2}}{7}$

Therefore: . $a = 4,\;\;\rlap{/////}b = 1 \;\;b = 2$

Edit: Right, masters! . . . b = 2 . . . Boy, that was a dumb error!

**masters** · August 20th 2010, 11:35 AM

$b=2$

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