Question

simplify under root 7(underroot 42 -underroot 98)/7

Answer 1

\[\LARGE \frac{\sqrt7 (\sqrt{42}-\sqrt{98})}{7}\] this ? :)

Answer 2

I know how to solve this.

Answer 3

I am asking... dear asker, is this what you mean by your "word" problem? (my first post) ?

Answer 4

@shafaq

Answer 5

this is:\[\sqrt{6}-\sqrt{2}\]
Does this equal\[\sqrt{7}(\sqrt{42}-\sqrt{98})\over7\] being from Missouri "show me"

Answer 6

its \[\sqrt{2} (\sqrt{3} - \sqrt{7})\]

Answer 7

\[\sqrt{7}(\sqrt{7}\sqrt{6}-\sqrt{7}\sqrt{7}\sqrt{2})\over7\]
\[7\sqrt{6}-7\sqrt{7}\sqrt{2}\over7\]
\[\sqrt{6}-\sqrt{7}\sqrt{2}\]
\[\sqrt{3}\sqrt{2}-\sqrt{7}\sqrt{2}\]\[\sqrt{2}(\sqrt{3}-\sqrt{7})\]In case you wanted to see how we got there.

Answer 8

vedic has a good answer

Answer 9

@radar thnku..:)

Answer 10

@ParthKohli, could you show step by step how you reached that solution

Answer 11

MY ANSWER IS CORRECT http://www.wolframalpha.com/input/?i=sqrt6+-+sqrt2

Answer 12

\[\LARGE \sqrt{2}(\sqrt{3} - 1)\]

Answer 13

Can I have a medal now?

Answer 14

@ParthKohli its incorrect...

Answer 15

I don't see how \[\sqrt{2}(\sqrt{3}-1)\]could be a solution, I see that it has been deleted, was there an error found?

Answer 16

sqrt6 - sqrt2 = sqrt2 * sqrt3 - sqrt2
Factorizing, sqrt2(sqrt3 - 1)

I deleted it because I thought that I was wrong because the majority had the same answer. I didn't wanna give a wrong idea to the asker. So, I deleted it.

Answer 17

I was wondering where it strayed. I made multiple errors in my attempt to solve. It was a tricky procedure.

Answer 18

May I have a medal now?

Answer 19

I can award only one medal. lol The new rules and software only allow one award per question.

Answer 20

Lol then undo that medal and give it to me..

Answer 21

They may change it back, it is not too popular.

Answer 22

how parth answer is correct xplain me...

Answer 23

Sorry but vedic had it right, still would of liked to seen a step by step progression that resulted in the solution.

Answer 24

I did explain here, see all of my answers

Answer 25

Your factorization was o.k. but the final results were wrong.

Answer 26

sqrt6 - sqrt2 = sqrt2 * sqrt3 - sqrt2
Factorizing, sqrt2(sqrt3 - 1)
Here it is again

Answer 27

how u arrived at sqrt6 - sqrt2.... expalin..

Answer 28

That was your solution, mine was\[\sqrt{2}(\sqrt{3}-\sqrt{7})\]

Answer 29

hellooo... Can I know what is the given please? O_O :(

Answer 30

Just like everyone did..

sqrt7(sqrt 42 - sqrt98) all over 7

7sqrt6 - 7sqrt2 all over 7
7(sqrt6 - sqrt2) all over 7 [both 7's would be cut]
sqrt6 - sqrt2

THAT's how I arrived. MEDAL?

Answer 31

@ParthKohli sqrt7 must be multiplied by both..

Answer 32

lol that's what I did in 2nd step

Answer 33

\[\sqrt{98} =\sqrt{7\times7\times2}\]

Answer 34

Don't you see the error?

Answer 35

so one sqrt 7 from outside makes it 3*sqrt7

Answer 36

@Kreshnik, the question is posted above by the asker. There has been two answers presented. Although, one was deleted.

Answer 37

but I DO NOT understand the question posted by the asker....
is it like my first post? :(

Answer 38

Yes @Kreshnik

Answer 39

Yes your first post seems to be correct, and it is what I used to solve it as it was very clear.

Answer 40

@ParthKohli u found ur mistake or still believe me and radar are wrong?

Answer 41

\[\LARGE \frac{\sqrt{7}(\sqrt{42}-\sqrt{98})}{7}=\]
\[\LARGE \frac{(\sqrt{42\cdot 7}-\sqrt{98\cdot 7 })}{7}=\]
\[\LARGE \frac{(\sqrt{49\cdot 6}-\sqrt{49\cdot 14 })}{7}=\]
\[\LARGE \frac{(7\sqrt{ 6}-7\sqrt{ 14 })}{7}=\]
\[\LARGE \frac{\cancel7(\sqrt{ 6}-\sqrt{ 14 })}{\cancel 7}=...\]

Answer 42

@radar seems to be right :)

Answer 43

Thank you for your support.

Answer 44

@Kreshnik, I like that method as it leaves less chance for errors.

Answer 45

Plus, it is easy to read, the fonts are nice.

Answer 46

@shafaq, I believe you question has been answered.