Question

Help needed! Big time.

Answer 1

with?

Answer 2

\[\frac{\sqrt{4y+2}+13}{6} =2\]

Answer 3

all izz well !

Answer 4

y = -1/4. Now someone verify it for me.

Answer 5

whats your progress/attempt ?

Answer 6

Show steps.

Answer 7

no, y=-1/4 isn't correct, 14/6 is not 2

Answer 8

@dmezzullo : http://1.bp.blogspot.com/-YD8Na5Mv4oY/USZJ0T5RKQI/AAAAAAAADnU/5U871_OaqRE/s1600/Ain-t-Nobody-Got-Time-Fo-Dat-sweet-brown-31241125-480-330.jpg

Answer 9

i am gettng no solution exist :O

Answer 10

Unless it be imaginary matey :D

Answer 11

when you squared both sides, you attached an extra root y=-1/4

Answer 12

Where?

Answer 13

yeah, answer to this is imaginary, or if we talk in real number terms, there's no solution

Answer 14

(root (...))^2 = (-1)^2

Answer 15

obviously 12-13=-1 .. that would mean you are going for imaginary !!

Answer 16

Wait, not imaginary... more like nonexistent, as @AravindG said...
I'm missing something, but I'm sure
\[\sqrt{x}=-1\]
has no solution

Answer 17

I'm confused.

Answer 18

i is defined as \[i^2=-1\]

Answer 19

Don't we always consider the positive square root?

Answer 20

I mean, sure, i is a solution to
\[\large x^2=-1\]
but
\[\large \sqrt{x}=-1\]
?

Answer 21

@terenzreignz yes but we can get imaginary number for positive root also

Answer 22

The answer is {} though

Answer 23

\[\sqrt{x}=i\]

Answer 24

oh , yes, here there no solution, not even imaginary.
from \(\sqrt{4y+2}=-1\)
you can directly say no solution or {}

Answer 25

as i said in my first statement

Answer 26

{} --> no solution exists.

Answer 27

BUT when we verify:

[sqrt(4*-1/4 + 2) + 13 ]/6
[sqrt(-1 + 2) + 13 ]/6
[sqrt(1) + 13 ]/6
14/6
7/3

Answer 28

wolf agrees
http://www.wolframalpha.com/input/?i=%28sqrt%284y%2B2%29%2B13%29%2F6%3D2

Answer 29

wolf does consider complex numbers right?
So even in Complex Numbers, no solution exists?

Answer 30

yeah, its 7/3, not 2

Answer 31

\[\dfrac{7}{3} \neq 2\]

Answer 32

yes, no solution even in complex numbers

Answer 33

so no solution

Answer 34

[sqrt(4*-1/4 + 2) + 13 ]/6
[sqrt(-1 + 2) + 13 ]/6
[sqrt(1) + 13 ]/6
14/6 or (-1+13)/6 < this bothers me.
7/3

Answer 35

sqrt 1 is never -1

Answer 36

Because we always consider the square root to be the principal square root, or else the square root is not a well-defined function.

Answer 37

yaa !!

Answer 38

you are considering positive square root

Answer 39

Why so?
When sqrt9 = +-3, so why can't this be?

Answer 40

no, sqrt 9 is just 3

Answer 41

noo sqrt 9 not equal to +-3

Answer 42

D:

Answer 43

square roots have only one root

Answer 44

Then what thing have 2 roots?

Answer 45

quadratic equations have 2 roots

Answer 46

Uh'oh a philosophical discussion coming up... :D
While it is true that +/- 3 is a solution to the equation
\[\large x^2 - 9 = 0\]
only 3 is a solution to the equation
\[\large x=\sqrt{9}\]

Answer 47

^ yes

Answer 48

Sorry, it makes more sense if written as
\[\large x^2 = 9\]

Answer 49

http://answers.yahoo.com/question/index?qid=20070907081106AAz7ys4

Answer 50

9 has two square roots a positive square root and a negative square root

when we usually speak of square root we talk about positive square root ie

\[\sqrt{9}, -\sqrt{9}\] are square root of 9

Answer 51

see that always a quantity under root is positive

Answer 52

Why me no get it?

Answer 53

because it is a common misconception from childhood !!

Answer 54

believe me i had the same problem 2 years before!

Answer 55

So we will just have one answer?

Answer 56

I mean 1 solution?

Answer 57

Oh, I seem to recall my teacher telling us that a way to define the absolute value for real numbers...
\[\large |x|=\sqrt{x^2}\]

Answer 58

Which would not make sense if the square root could be negative.

Answer 59

exactly !!!

Answer 60

really worth checking out :: http://en.wikipedia.org/wiki/Square_root

Answer 61

unless specified we always talk of principal square root

Answer 62

Hmm. I guess i get it.
Thanks @AravindG @hartnn @terenzreignz

Answer 63

be careful next time :)

Answer 64

*Hope so* :D

Answer 65

as @terenzreignz stated :)

Answer 66

:)
\[\huge \sqrt{x} \ \ge \ 0\]

Answer 67

its actually \(\sqrt{x^2}=|x|\)
and not \(\sqrt{x}=|x|\)

Answer 68

Lol.

Answer 69

oopsie i made a blunder there ..thanks for pointing out

Answer 70

it means this question is wrong

Answer 71

lol

Answer 72

Haha I love my neat work :3