## anonymous one year ago Which one is greater? =)

1. anonymous

$\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 3 }}$ or $\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}$

2. anonymous

i got the second one as bigger

3. ganeshie8

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

4. ganeshie8

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

5. anonymous

im on a diet so 1/4

6. anonymous

but 1/3 is bigger lol

7. ganeshie8

so...

8. anonymous

wait but can u take a look at what i did

9. anonymous

|dw:1437674327047:dw|

10. ganeshie8

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

11. anonymous

but doesnt it become a 1?

12. ganeshie8

nope

13. anonymous

what does it become then?

14. ganeshie8

|dw:1437674587867:dw|

15. anonymous

then whats the proper way to do it?

16. 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 $$\frac{ 1 }{ \sqrt{1} }+\sqrt{\frac{ 1 }{ 4 }}$$

17. anonymous

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

18. anonymous

like how do the solutions compare?

19. ganeshie8

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

20. ganeshie8

@dan815

21. anonymous

ok can u help me with another problem then?

22. anonymous

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

23. 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

24. anonymous

25. anonymous

26. anonymous

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

27. freckles

@ganeshie8 oh how to compare those two numbers?

28. 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....

29. anonymous

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

30. 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 $1+\sqrt{\frac{1}{3}} \text{ ? } 1 +\sqrt{\frac{1}{4}} \\ \text{ subtract 1 on both sides } \sqrt{\frac{1}{3}} ? \sqrt{\frac{1}{4}} \\ \text{ squaring both sides } \frac{1}{3} ? \frac{1}{4} \\ \\ \text{ well we know } 4>3 \text{ so } \frac{1}{4} < \frac{1}{3} \\ \text{ which means } \sqrt{\frac{1}{4}} < \sqrt{\frac{1}{3}} \\ \text{ which means } 1+\sqrt{\frac{1}{4}} < 1 +\sqrt{\frac{1}{3}}$

31. anonymous

hmmmm okay..

32. anonymous

that still wasn't solved tho =(

33. freckles

what do you mean ?

34. freckles

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

35. anonymous

like the problem solved out

36. freckles

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

37. anonymous

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

38. anonymous

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

39. anonymous

don't*

40. freckles

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41. freckles

so you are right it is 2 or -8

42. anonymous

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

43. anonymous

oh okay but why is it positive 8 then?

44. freckles

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

45. 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

46. anonymous

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

47. freckles

no*

48. freckles

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

49. freckles

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

50. anonymous

yea

51. anonymous

so positive 8 is wrong?

52. 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

53. anonymous

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

54. freckles

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

55. 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

56. anonymous

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57. anonymous

like i don't understand why thats wrong

58. freckles

these look different then the other two values you gave

59. 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 ?

60. anonymous

|dw:1437676741343:dw|

61. anonymous

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

62. 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) $\frac{\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}} \\ \frac{\sqrt{3}+1}{\sqrt{3}} \text{ but you cannot say } \frac{a+b}{a} =\frac{\cancel{a}+b}{\cancel{a}}=1+b \\ \text{ but you can say} \frac{a+b}{a}=\frac{a}{a}+\frac{b}{a}=1+\frac{b}{a}$

63. freckles

you can "cancel out" common factors in division

64. freckles

not common terms

65. anonymous

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

66. freckles

fraction bar is division

67. 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

68. freckles

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

69. anonymous

Like for this one |dw:1437677209526:dw|

70. 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 $\frac{6x^2+5x+1}{2x-1}$

71. anonymous

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

72. freckles

nope

73. freckles

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

74. anonymous

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

75. anonymous

|dw:1437677555931:dw|

76. freckles

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

77. freckles

you could but another way is the way I did

78. freckles

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

79. anonymous

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

80. anonymous

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

81. anonymous

SORRY hahahaha okay i got it lmaoooooo

82. anonymous

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

83. anonymous

ahhhh i see lolol

84. anonymous

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

85. freckles

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

86. 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

87. freckles

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

88. freckles

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

89. anonymous

lol gotcha <3

90. 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

91. anonymous

kk

92. freckles

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

93. anonymous

thank u again =)

94. freckles

np