how to no when square root of x is positive or negative (1 Viewer)

[totallybord]

ok i no that the square root of x is by definition positive rite?
and negative square root of x is negative
but when am i spose to no that in some questions like for completing the square stuff theres 2 answers and other stuff theres only one and its soo freakin confusing
helpp pleaseee

 

[shinji]

erm.. not really getting ur question.

when u squareroot x, u get 2 answers. However, if ur in a restricted domain for x, the answer that is outside of the domain is wrong.

i think that's what u meAN?

 

[blakwidow]

square root of x = positive and negative answer

for example square root 4 = +2 or -2 because if you square both you can get 4

BUT

cube root of 8 = positive answer only

-------------------------

Post a question up so we can help further

-blakwidow

 

[watatank]

square root of x = positive and negative answer

for example square root 4 = +2 or -2

no its not. square root of any real number is positive.

square root of 4 is 2. period.

however if x² = 4 then

x = +/- sqrt4

x = +/- 2

uhm in answer to the original poster, an example would be handy

 

[bos1234]

no its not. square root of any real number is positive.

square root of 4 is 2. period.

however if x² = 4 then

x = +/- sqrt4

x = +/- 2

uhm in answer to the original poster, an example would be handy

but -2 x -2 = 4 as well?

 

[watatank]

saying a number squared is 4 is not the opposite of the square root of 4.

i never did four unit but in the maths you learn in two and three unit when you take a square root of 4 you only take the positive case. when you say a number squared is 4 you have to take the positive case and the negative case

 

[pLuvia]

If you are completing the square let's say you are using that method and you get down to this equation
(ax+b)²=c, where a,b,c are real numbers
(ax+b)=+sqrt{c}
ax=-b+sqrt{c}
x=(-b+sqrt{c})/a

If you are completing the square you take both positive and negative answers unless there's a domain with a restriction

But if it's an equation like
x²=9
x=+3
You take the positive if the domain restricts it to the positive or vice versa

Is that what you mean?

 

[PC]

Basically, there's two ways you can go.

1. If you are taking the square root of both sides of an equation, then there are two solutions, one positive and one negative.

For example:
x² = 9
Take square root of both sides:
x = 3

Similarly, when completing the square to solve a quadratic:
x² + 6x – 1 = 0
x² + 6x = 1
Add the square of half the coefficient of x to both sides:
x² + 6x + 9 = 1 + 9
x² + 6x + 9 = 10
(x + 3)² = 10
Now take the square root of both sides:
x + 3 = √10
Subtract 3 from both sides:
x = –3 √10

The other way you can go is:

2. The square root symbol means, by definition, the positive square root only.

So that's why in the quadratic formula, you've got the sign in there. The root sign on its own only indicates one particular value, the positive root.

So ... try these two equations:

1. x² = 9
.: x = 3

or:

2. x = √9
.: x = 3

There is a very subtle difference!

 

[totallybord]

i get it kind of...
but x^2 = 9 is pretty much x= square root of 9 ....but has 2 answers...
how r u spose to differentiate btwn them?

 

[pLuvia]

If it's just down right solve x²=9, then the answer is + and - 3, if there is a domain let's say, solve x²=9 for x>0, then you only take the positive answer and vice versa if the domain is x<0

You'd better give us an example if that didn't make sense

 

[Forbidden.]

ok i no that the square root of x is by definition positive rite?
and negative square root of x is negative
but when am i spose to no that in some questions like for completing the square stuff theres 2 answers and other stuff theres only one and its soo freakin confusing
helpp pleaseee

Then ....

erm.. not really getting ur question.

when u squareroot x, u get 2 answers. However, if ur in a restricted domain for x, the answer that is outside of the domain is wrong.

i think that's what u meAN?

If it's just down right solve x²=9, then the answer is + and - 3, if there is a domain let's say, solve x²=9 for x>0, then you only take the positive answer and vice versa if the domain is x<0

You'd better give us an example if that didn't make sense

Domain: x = 0 & x > 0
Range: All real x > 0

for

y = √x

http://img407.imageshack.us/img407/7590/sqrtxsg3.jpg

So if x > 0 then y > 0 unless y = - √x

 

[brows]

no its not. square root of any real number is positive.

square root of 4 is 2. period.

however if x² = 4 then

x = +/- sqrt4

x = +/- 2

uhm in answer to the original poster, an example would be handy

I agree with u. U cant take the square root of a negative.

 

[Slidey]

I agree with u. U cant take the square root of a negative.

How much would you be willing to bet on that? Now, it's certainly not 2u or 3u, but you definitely do it in 4u.

It's like this:

sqrt(-9) is 3sqrt(-1) for clarification.

Courtesy of: http://xkcd.com/c179.html

 

[watatank]

I agree with u. U cant take the square root of a negative.

you can, but for the purposes of this discussion lets just say it doesnt (this is posted in the 2unit maths forum)

 

[totallybord]

ok thanks
umm could u help with some other questions?
what is the easiest way to determine domain and range of an equation
and also for y=x^2 +x+2 , when u draw th e thing, u draw 2 separate graphs and then u add the y coordinates together right? and for y=x^2 -x-2 u subtract the y coordinates rite? but for absolute values, such as y=|x^2| +|x+2| and y=|x^2| -|x-2| u always add them together right?
oh and also, if the question says, show that if -2(less than/equal to)x(less than/equal to)2, then G(x) = 4 where G(x)=|x+2|+|x-2|....
how do i show this ? by doing subbing all the nos into the equation? is that how u set it out?
thank u for all ur help

 

[ellen.louise]

That was my reaction...

 

[pLuvia]

ok thanks
umm could u help with some other questions?
what is the easiest way to determine domain and range of an equation
and also for y=x^2 +x+2 , when u draw th e thing, u draw 2 separate graphs and then u add the y coordinates together right? and for y=x^2 -x-2 u subtract the y coordinates rite? but for absolute values, such as y=|x^2| +|x+2| and y=|x^2| -|x-2| u always add them together right?
oh and also, if the question says, show that if -2(less than/equal to)x(less than/equal to)2, then G(x) = 4 where G(x)=|x+2|+|x-2|....
how do i show this ? by doing subbing all the nos into the equation? is that how u set it out?
thank u for all ur help

The domain is the whole if the x axis, but the range is from the y intercept to infinity.

Domain = {xER}
Range = {yER; y>2}

where E is "an element of" the little "e"

You could do it that method but why bother with all the addition of ordinates? You can easily find the y intercept, the stat point and by obeservation it can be seen it doesn't have any x intercepts.

For the next one

g(x)=|x+2|+|x-2|
For the domain of -2<x<2
|x+2|=x+2
|x-2|=-x+2
Hence g(x)=(x+2)+(-x+2)=4