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

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?

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

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 $a+b=18$.

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

\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, $a=15,b=3$.

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