- #1

- 76

- 0

(2x-1)sqrtof x-3x

is it just like normal distribution? Would I just put the answer underneath the square root?

sqrt2x^2-6x^2-x+3x?

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- Thread starter camel-man
- Start date

- #1

- 76

- 0

(2x-1)sqrtof x-3x

is it just like normal distribution? Would I just put the answer underneath the square root?

sqrt2x^2-6x^2-x+3x?

- #2

- 81

- 0

all of these things are equal:

a*sqrt(b) = sqrt(a*a) * sqrt(b) = sqrt(a*a*b)

a*sqrt(b) = sqrt(a*a) * sqrt(b) = sqrt(a*a*b)

- #3

- 795

- 7

(2x-1)sqrtof x-3x

is it just like normal distribution? Would I just put the answer underneath the square root?

sqrt2x^2-6x^2-x+3x?

To figure something like this out, try it with regular old numbers.

For example [itex]5 \sqrt{4} = 5 * 2 = 10[/itex].

But if you just put the 5 under the square root sign to make it sqrt(5*2) = sqrt(10), then that's not the same thing as 10 so you can't do that.

Why not? Well, [itex]\sqrt{a^2b^2} = ab[/itex], right? That's because

(ab)

So, what's the fix? If we have 5 * sqrt(4) we can put the 5 under the radical by

[itex]5 \sqrt{4} = \sqrt{5^2*4} = \sqrt{100} = 10[/itex] as it should be.

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- #4

- 76

- 0

i just square 9y+1 and put it under the square root with 82 and then times them both together?

- #5

- 795

- 7

i just square 9y+1 and put it under the square root with 82 and then times them both together?

Yes, but now you have to be careful. If 9y+1 is negative, squaring it will lose information. So this depends on the context.

In other words it is not always true that [itex]\sqrt{x^2} = x[/itex]. That's because the meaning of the square root symbol is the

Why do you want to put this expression under the radical? In general, doing so will change the meaning and introduce an error.

- #6

- 76

- 0

A= 2pi integral from a to b f(x)sqrt 1+ f(x) prime^2

that is the forumula that I have to use

- #7

HallsofIvy

Science Advisor

Homework Helper

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[tex]\int_a^b f(x)\sqrt{1+ f'^2(x)}dx[/tex]

Yes, you

[tex]\int_a^b \sqrt{f^2(x)(1+ f'^2(x))}dx[/tex]

Whether that is a good idea or not depends upon f.

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