Which one is greater? =)

- anonymous

Which one is greater? =)

- jamiebookeater

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- anonymous

\[\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }}\] or
\[\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}\]

- anonymous

i got the second one as bigger

- ganeshie8

which one is big, 1/3 or 1/4 ?

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## More answers

- ganeshie8

would you like to have 1/3 of cheesecake
or
1/4 of cheesecake ?

- anonymous

im on a diet so 1/4

- anonymous

but 1/3 is bigger lol

- ganeshie8

so...

- anonymous

wait but can u take a look at what i did

- anonymous

|dw:1437674327047:dw|

- ganeshie8

we can't cancel like that
\[\dfrac{a+b}{b}~~\ne~~a\]

- anonymous

but doesnt it become a 1?

- ganeshie8

nope

- anonymous

what does it become then?

- ganeshie8

|dw:1437674587867:dw|

- anonymous

then whats the proper way to do it?

- ganeshie8

you have said earlier that 1/3 is greater than 1/4
therefore
\(\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }}\) is greater than \(\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}\)

- anonymous

but can you solve it out? I need to know why....

- anonymous

like how do the solutions compare?

- ganeshie8

im not getting any ideas, not seeing any neat/standard way for problems like these @freckles

- ganeshie8

- anonymous

ok can u help me with another problem then?

- anonymous

it's an easy question but i don't know why this website is giving a wrong answer....

- anonymous

If \[(x+3)^{2}=25\] which of the following could be the value of x?
a. -8
b. -5
c. -2
d. 5
e. 8

- anonymous

i thought it was this at first |dw:1437675183398:dw|

- anonymous

then i thought it was this : |dw:1437675254689:dw|

- anonymous

but the correct answer is 8... and i don't know why

- freckles

@ganeshie8 oh how to compare those two numbers?

- anonymous

Hi freckles for the first problem I presented can you solve it out for me? because i know that 1/3 is larger but i need to know how the problem is completely solved and compare the two solutions....

- anonymous

I gues i did it the wrong way but fail to understand why it's wrong

- freckles

first of all you really did mean:
\[
\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }}
\text{ or }
\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}
\]
just asking because I find that they put sqrt(1) weird
you know because that is 1
and also find it weird they put 1/sqrt(1) because that is also 1
\[1+\sqrt{\frac{1}{3}} \text{ ? } 1 +\sqrt{\frac{1}{4}} \\ \text{ subtract 1 on both sides } \sqrt{\frac{1}{3}} ? \sqrt{\frac{1}{4}} \\ \text{ squaring both sides } \frac{1}{3} ? \frac{1}{4} \\ \\ \text{ well we know } 4>3 \text{ so } \frac{1}{4} < \frac{1}{3} \\ \text{ which means } \sqrt{\frac{1}{4}} < \sqrt{\frac{1}{3}} \\ \text{ which means } 1+\sqrt{\frac{1}{4}} < 1 +\sqrt{\frac{1}{3}}\]

- anonymous

hmmmm okay..

- anonymous

that still wasn't solved tho =(

- freckles

what do you mean ?

- freckles

i guess I'm asking what are you looking for?

- anonymous

like the problem solved out

- freckles

to me I showed why sqrt(1/4)3 which means 1/4<1/3
taking square root of both sides the direction of inequality still holds

- anonymous

hmmm okay i gotcha can you help me with the second problem please?

- anonymous

i just really fon't understand why it's 8 instead of 2 or -8

- anonymous

don't*

- freckles

|dw:1437675933396:dw|

- freckles

so you are right it is 2 or -8

- anonymous

wait what? is this for all square root problems? it's either a 5 or -5?

- anonymous

oh okay but why is it positive 8 then?

- freckles

\[x^2=a \text{ if } a \text{ is positive then yes } x=\sqrt{a} \text{ or } x=-\sqrt{a} \\ \text{ since both } (\sqrt{a})^2=a \text{ and } (-\sqrt{a})^2=a\]

- freckles

I guess I could also include 0 and negative a
but x^2=0 only gives x=0
and x^2=a where a is negative gives complex solutions

- anonymous

so you're saying it can be 2 and -2 and also 8 and -8?

- freckles

no*

- freckles

\[(x+3)^2=25 \text{ gives us } x+3=5 \text{ or } x+3=-5 \]
is what I'm saying

- freckles

\[\text{ which gives } x=5-3 \text{ or } x=-5-3 \]

- anonymous

yea

- anonymous

so positive 8 is wrong?

- freckles

yeah if the problem really is (x+3)^2=25
because if x=8 then (8+3)^2=(11)^2=121
121 is not 25

- anonymous

ahhhhh okay i gotcha ^_^ thank youuuuuuuuuuuuuu!!~~~

- freckles

can we go back to the previous problem @yomamabf
were you expecting a certain method on that one?

- anonymous

Well i wanted to have it solved out so i know that the end solution would coincide with the rule that the smaller denominator will always produce a larger number

- anonymous

|dw:1437676640564:dw|

- anonymous

like i don't understand why thats wrong

- freckles

these look different then the other two values you gave

- freckles

\[\frac{1}{\sqrt{1}}+\sqrt{\frac{1}{4}} \text{ or } \frac{1}{\sqrt{1}}+\sqrt{\frac{1}{3}} ?\]
are these not the two values you were comparing ?

- anonymous

|dw:1437676741343:dw|

- anonymous

yea those are but i just solved it is why they look different

- freckles

\[\frac{1}{\sqrt{1}}+\frac{\sqrt{1}}{\sqrt{3}} \\ \]
So you tried to combine fractions :
This means you want to multiply the first fraction by sqrt(3)/sqrt(3)
\[\frac{\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}} \\ \frac{\sqrt{3}+1}{\sqrt{3}} \text{ but you cannot say } \frac{a+b}{a} =\frac{\cancel{a}+b}{\cancel{a}}=1+b \\ \text{ but you can say} \frac{a+b}{a}=\frac{a}{a}+\frac{b}{a}=1+\frac{b}{a}\]

- freckles

you can "cancel out" common factors in division

- freckles

not common terms

- anonymous

oh okay so u can cancel out if you divide but cant reduce in this situation because we're not dividing?

- freckles

fraction bar is division

- anonymous

so when can i cancel them out ? because I remember doing that before for another problem and it was the right thing to do.... =( so confused

- freckles

\[\text{ say you have this } \\ \frac{\sqrt{3}+2 \sqrt{3}}{\sqrt{3}} \\ \text{ the \top and bottom share a common factor } \sqrt{3} \\ \ \text{ that is you can factor out a } \sqrt{3} \text{ on \top } \\ \frac{\sqrt{3}(1+2)}{\sqrt{3}} \\ \text{ notice top and bottom have } \sqrt{3} \text{ factor \in common } \\ \text{ we know } \frac{\sqrt{3}}{\sqrt{3}} =1 \text{ so we can do } \\ \frac{\cancel{\sqrt{3}}(2+1)}{\cancel{3}}=\frac{2+1}{1}=2+1=3\]

- anonymous

Like for this one |dw:1437677209526:dw|

- freckles

basically all terms have the have the same factor
for example:
\[\frac{6x^3+5x^2+x}{2x^2-x}\]
all the terms on top have a common factor x
all terms on bottom have a common factor x
you can divide each term by x
\[\frac{6x^2+5x+1}{2x-1}\]

- anonymous

yea i understand that but for roots its a bit different right?

- freckles

nope

- freckles

\[\frac{ 6 (\sqrt{3})^3+5 (\sqrt{3})^2+\sqrt{3}}{2 (\sqrt{3})^2-\sqrt{3}} \text{ divide top and bottom by } \sqrt{3} \\ \text{ divide each term in the top by } \sqrt{3} \\ \text{ divide each term in the bottom by } \sqrt{3} \\ \frac{6 (\sqrt{3}))^2+5 \sqrt{3}+1}{2 \sqrt{3}-1}\]

- anonymous

hey for the problem u put on top aren't u supposed to solve it like this? |dw:1437677469609:dw|

- anonymous

|dw:1437677555931:dw|

- freckles

\[\frac{\sqrt{3}+1}{\sqrt{3}} \\ \text{ if you did \choose to try what you did earlier } \\ \text{ you would also need to divide the } 1 \text{ by } \sqrt{3} \\ \text{ that is you would have } \\ \frac{\frac{\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{ \frac{\sqrt{3}}{\sqrt{3}}} =\frac{1+\frac{1}{\sqrt{3}}}{1}=1+\frac{1}{\sqrt{3}} \text{ but this is what you had to \begin with } \\ \text{ so you were working backwards to what you started with }\]

- freckles

you could but another way is the way I did

- freckles

and the way I did it was to give you an example of when you can cancel

- anonymous

if it involves addition and subtraction don't reduce it but if it involves multiplication ad division then reduce it

- anonymous

oh wait nevermind HAHAHAHAHHAHA OHHHHHHHHHHHHHHHHHHH IM SO STUPID OKAY I GOT IT

- anonymous

SORRY hahahaha okay i got it lmaoooooo

- anonymous

i see what i did wrong this is what it's supposed to be |dw:1437677816464:dw|

- anonymous

ahhhh i see lolol

- anonymous

thank you so much for breaking everything down you're the best <33333

- freckles

no problem so
you can do:
\[\frac{5+10}{5}=\frac{5(1+2)}{5}=1+2=3 \text{ but you can't do } \frac{\cancel{5}+10}{\cancel{5}}=1+10=11 \\ \text{ and yes another way \to do } \frac{5+10}{5} \text{ is by adding the numerator first } \frac{15}{5}=3\]

- anonymous

ahhhhh i see always have to simplify everything on top in the numerator you can't just do 1 and not do the rest... i see now lol thank youuuuuu

- freckles

right everything on top has to be divided by the thing on bottom

- freckles

like we can't do part of the top divided by the bottom and just ignore the rest being dividing by the bottom

- anonymous

lol gotcha <3

- freckles

another way to look at (5+10)/5
is looking at it as 5/5+10/5
separating the fraction first
doing the divisions
1+2
then doing the additions

- anonymous

kk

- freckles

anyways I think you really do get it now (I think :p)
so I will leave you alone now :p

- anonymous

thank u again =)

- freckles

np

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