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I am stuck at a question.

If $\sqrt{18-6\sqrt{5}} = \sqrt{a}- \sqrt{b}$, then which of the following is true:

  1. $a+b= 18$
  2. $a+b= 16$
  3. $a+b= 20$
  4. $a-b= 18$

I tried to first take the main root to the other side which made the R.H.S $2\sqrt{ab}$ but after that, I can't solve this any further.

Can someone help me in telling me what to do next?

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    $\begingroup$ Are $a,b \in \mathbb{Q}$? $\endgroup$ – Zardo Jun 19 '15 at 12:39
  • $\begingroup$ I am just a IX grade student, I don't understand that. $\endgroup$ – codetalker Jun 19 '15 at 12:40
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Consider

$$(\sqrt a-\sqrt b)^2=a+b-2\sqrt{ab}=18-6\sqrt5=18-2\sqrt{45}$$

For integer $a,b$, integral part of the expression is 18.

So $a+b=18$.

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  • $\begingroup$ Thanks but how can it be $18+2√45$ when it is $18 - 6√5$? $\endgroup$ – codetalker Jun 19 '15 at 12:51
  • $\begingroup$ and also we must use $(a-b)^2$ $\endgroup$ – codetalker Jun 19 '15 at 12:53
  • $\begingroup$ Yah some typo. Fixed $\endgroup$ – Mythomorphic Jun 19 '15 at 12:54
  • $\begingroup$ btw $-6\sqrt5=-2\sqrt{5\cdot3^2}=-2\sqrt{45}$ $\endgroup$ – Mythomorphic Jun 19 '15 at 12:55
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$$\begin{align}\sqrt{18-6\sqrt 5}&=\sqrt{18-2\sqrt{5\times 3^2}}\\&=\sqrt{(15+3)-2\sqrt{15\times 3}}\\&=\sqrt{(\sqrt{15}-\sqrt{3})^2}\\&=\sqrt{15}-\sqrt 3\end{align}$$ So, $a=15,b=3$.

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  • $\begingroup$ Yeah let me try that.... Thanks @mathlove $\endgroup$ – codetalker Jun 19 '15 at 12:43

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