Vantage

New member
I'm working on A-Level maths indices problems, and I encountered this question:

"a) Write [imath]\sqrt{27} + \sqrt{3}[/imath] in the form [imath]\sqrt{a}[/imath]." - This I had no issue with. [imath]\sqrt{48}[/imath] was my answer.

Followed by:

"b) Without using decimal approximations, explain whether [imath]\sqrt{27} - \sqrt{20}[/imath] is greater or less than [imath]\sqrt{5} - \sqrt{3}[/imath]."

How would I go about doing this? My attempts haven't worked at all, I'd appreciate any help.

Thanks in advance. skeeter

Elite Member
start by simplifying ...

$\sqrt{27}-\sqrt{20} = 3\sqrt{3} - 2\sqrt{5}$

Vantage

New member
start by simplifying ...

$\sqrt{27}-\sqrt{20} = 3\sqrt{3} - 2\sqrt{5}$
I got this far, then I got stumped. How do I find which is greater without knowing decimal approximations?

Otis

Elite Member
I got [that] far, then I got stumped.
Hi Vantage. Next time, please share how far you got when you start a thread. That way, we'll know where you're stuck (saves you time).

Now pick one of the two possible cases, for example:

[imath]\sqrt{27} - \sqrt{20}\; \text{ is less than } \;\sqrt{5} - \sqrt{3}[/imath]

Write and rearrange that inequality (i.e., first separate and then combine like-terms), and use what you discovered in part (a). Vantage

New member
Hi Vantage. Next time, please share how far you got when you start a thread. That way, we'll know where you're stuck (saves you time).
My apologies. I'll be sure to do that next time I post. Write and rearrange that inequality (i.e., first separate and then combine like-terms), and use what you discovered in part (a). Thank you for this. This works perfectly, I just didn't put 2 and 2 together... • Otis

Jomo

Elite Member
I would start off with
5 = sqrt(25) < sqrt(27) < sqrt(36) = 6. So 5 < sqrt(27) < 6
4 = sqrt(16) < sqrt(20) < sqrt(25) = 5. So 4 < sqrt(20) < 5 or -5 < -sqrt(20) < -4

Adding yields that 0 < sqrt(27) - sqrt(20) < 2

Do the same with the other expression.

This is what I would try 1st. It actually doesn't help, but not all attempts to solving a problem works!

Vantage

New member
I would start off with
5 = sqrt(25) < sqrt(27) < sqrt(36) = 6. So 5 < sqrt(27) < 6
4 = sqrt(16) < sqrt(20) < sqrt(25) = 5. So 4 < sqrt(20) < 5 or -5 < -sqrt(20) < -4

Adding yields that 0 < sqrt(27) - sqrt(20) < 2

Do the same with the other expression.

This is what I would try 1st. It actually doesn't help, but not all attempts to solving a problem works!
In my attempts, I did something similar to this. I found the upper and lower bounds of [imath]\sqrt{27}, \sqrt{20}, \sqrt{5}, \sqrt{3},[/imath] in terms of squares of integers. Then I tried the calculations, but using the bounds, the calculations just ended up being equal in both scenarios. So in the end it didn't work but it did help me understand the question a little better. skeeter

Elite Member
If [imath]\sqrt{27} - \sqrt{20} > \sqrt{5} - \sqrt{3}[/imath] ...

[imath](3\sqrt{3} - 2\sqrt{5}) - (\sqrt{5} - \sqrt{3}) > 0[/imath]

[imath]4\sqrt{3} - 3\sqrt{5} > 0[/imath]

[imath]4\sqrt{3} >3\sqrt{5}[/imath]

multiply both sides by [imath]\sqrt{3}[/imath] ...

[imath]4 \cdot 3 > 3\sqrt{15}[/imath]

[imath]\sqrt{16} \cdot 3 > 3 \cdot \sqrt{15} \implies \sqrt{16} > \sqrt{15}[/imath], a true statement.

• Otis and lookagain