Surds

[Probability]

I have a Surd and am asked to multiply it out.

[MATH]\sqrt{7}\times\sqrt{14}[/MATH]
This is my line of reasoning but I'm questioning my reasoning getting to the solution. I'm not confident I'm using the correct procedure!

Am I mathematically thinking this out with the numbers the using approved methods!

[MATH]\sqrt{7}\times\sqrt{14}=\sqrt{{7\times 14}}=\sqrt{98}[/MATH]
From this point I'm looking for the largest number that will go into 98. I find that 14 can be taken from 98 seven times, and 7 taken from 14 twice, hence;

[MATH]\sqrt{7}\times\sqrt{14}=\sqrt{({7}\times{14})}=\sqrt{98}=\sqrt{({7^2\times{2}})}=7\sqrt{2}[/MATH]

 

[Albi]

The last part is not correct the root of 98 is not equal to the root of 7×2 and the root of 7×2 is not 7×the root of 2 (the post above #1 has been corrected - there were some syntax error - no major math-error)

 

[Probability]

The last part is not correct the root of 98 is not equal to the root of 7×2 and the root of 7×2 is not 7×the root of 2

That's interesting. I'm also learning this as I go along so any advice is very much appreciated, but could you please explain why they are not equal?

 

[pka]

[MATH]\sqrt{7}\times\sqrt{14}[/MATH]This is my line of reasoning but I'm questioning my reasoning getting to the solution. I'm not confident I'm using the correct procedure!

That is correct but I would do this:
\(\sqrt{7}\cdot\sqrt{14}=\sqrt{7}\cdot(\sqrt{7}\cdot\sqrt{2})=7\sqrt2\).

 

[Probability]

That is correct but I would do this:
\(\sqrt{7}\cdot\sqrt{14}=\sqrt{7}\cdot(\sqrt{7}\cdot\sqrt{2})=7\sqrt2\).

Thanks I'm still learning the latex as I go

 

[Albi]

q

Thanks I'm still learning the latex as I go
[/QUOTE

That is correct but I would do this:
\(\sqrt{7}\cdot\sqrt{14}=\sqrt{7}\cdot(\sqrt{7}\cdot\sqrt{2})=7\sqrt2\).
[/Q

That is correct but I would do this:
\(\sqrt{7}\cdot\sqrt{14}=\sqrt{7}\cdot(\sqrt{7}\cdot\sqrt{2})=7\sqrt2\).

I would remove the brackets completely

 

[Probability]

Trying to continue building my confidence with these Surds.

When asked to multiply roots, would this be a good method;

[MATH]2\sqrt3\cdot2\sqrt15=2\cdot2\sqrt3\cdot\sqrt15=4\sqrt3\cdot\sqrt15=[/MATH]
[MATH]4\sqrt45=12\sqrt5[/MATH]
Is the surd now complete at this point or should I continue as follows;

[MATH]4\sqrt(9\cdot5)=[/MATH]
[MATH]4\sqrt3\sqrt5=12\sqrt5[/MATH]

 

[lookagain]

Trying to continue building my confidence with these Surds.
.
.
[MATH]4\sqrt(9\cdot5)=[/MATH]
[MATH]4\sqrt3\sqrt5=12\sqrt5 \ \ \ \ [/MATH] false

Regarding your earlier lines, break down your medium
size radicals so that you do not have to undo your
work and have to break down even bigger radicals:

\(\displaystyle 2\sqrt{3}\cdot2\sqrt{15} \ = \ \)

\(\displaystyle 2\sqrt{3}\cdot2\sqrt{3}\cdot\sqrt{5} \ = \)

\(\displaystyle 2\cdot2\cdot3\cdot\sqrt{5} \ = \)

\(\displaystyle 12\sqrt{5}\)

 

[Probability]

Regarding your earlier lines, break down your medium
size radicals so that you do not have to undo your
work and have to break down even bigger radicals:

\(\displaystyle 2\sqrt{3}\cdot2\sqrt{15} \ = \ \)

\(\displaystyle 2\sqrt{3}\cdot2\sqrt{3}\cdot\sqrt{5} \ = \)

\(\displaystyle 2\cdot2\cdot3\cdot\sqrt{5} \ = \)

\(\displaystyle 12\sqrt{5}\)

I should not have put the last 3 in a radical should I. I did not understand that.

I should have wrote;

[MATH]4\cdot3\sqrt5=12\sqrt5[/MATH]

 

[lookagain]

I should not have put the last 3 in a radical should I. I did not understand that.

I should have wrote;

[MATH]4\cdot3\sqrt5=12\sqrt5[/MATH]

You are correct that the last 3 does not go inside the radical sign. Once
you take the square root of 9, the square root symbol is gone. You end
up with the integer 3 for that part of the step.

 

[Probability]

They say practice makes perfect. I got admit these surds for me are taking me some time to get my head round the understanding, even when I think I understand I make mistakes.

I've been struggling a little while getting this one completed.

[MATH]5\sqrt{8}-{2}\sqrt{2}=({5}-{2})={3}\sqrt{8}-\sqrt{2}=[/MATH]
[MATH]3\sqrt({4}\times{2})-\sqrt(2)=[/MATH]
[MATH]3\sqrt({2}\times2)-{2}\sqrt(2)=[/MATH]
[MATH]3\sqrt{4}-{2}\sqrt(2)=[/MATH]
[MATH]{12}-{2}\sqrt{2}[/MATH]
[MATH]{10}\sqrt{2}[/MATH]
Hopefully all my headaches and hard work might help others out.

 

[Probability]

Just checking my rough work it seems I've either made a mistake or not quite finished the surd above. It should finish with [MATH]{8}\sqrt{2}[/MATH]Looks like I'll need to look at it again

 

[Probability]

I can see where my understanding was incorrect now. What I did was to subtract 2 from 5 which was the wrong technique. What I should have done is subtract 5 from the LHS and add minus 2 to the RHS, hence I'd of ended with;

[MATH]{5}\sqrt{8}-{2}\sqrt{2}={5}\sqrt{8}-{2}\sqrt{2}=[/MATH]
[MATH]5\sqrt(4\times2)-{2}\sqrt{2}[/MATH]
[MATH]5\sqrt(2\times2)-{2}\sqrt{2}=[/MATH]
[MATH]10-2\sqrt{2}={8}\sqrt{2}[/MATH]

 

[Dr.Peterson]

How did 4 change to 2 with nothing else changing? And how does the order of operations apply to the last line?

This is the sort of question you have to ask yourself when you check whatever you write! What changed from one line or expression to the next, and why?

 

[Probability]

I can see where my understanding was incorrect now. What I did was to subtract 2 from 5 which was the wrong technique. What I should have done is subtract 5 from the LHS and add minus 2 to the RHS, hence I'd of ended with;

[MATH]{5}\sqrt{8}-{2}\sqrt{2}={5}\sqrt{8}-{2}\sqrt{2}=[/MATH]
[MATH]5\sqrt(4\times2)-{2}\sqrt{2}[/MATH]
[MATH]5\sqrt(2\times2)-{2}\sqrt{2}=[/MATH] ((This is where I think the mistake is, in this line))

[MATH]10-2\sqrt{2}={8}\sqrt{2}[/MATH]

[MATH]{5}\times{2}\sqrt{2}-{2}\sqrt{2}[/MATH]
[MATH]{10}\sqrt{2}-{2}\sqrt{2}={8}\sqrt{2}[/MATH]
Does this look any better Dr.Peterson