Distributing into a square root

[camel-man]

Its been a while since I have taken any kind of math class, I am a bit rusty in general algebra. Can someone explain how I would multiply an equation like this

(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?

 

[marty1]

all of these things are equal:

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

 

[SteveL27]

Its been a while since I have taken any kind of math class, I am a bit rusty in general algebra. Can someone explain how I would multiply an equation like this

(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 5 \sqrt{4} = 5 * 2 = 10.

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, \sqrt{a^2b^2} = ab, right? That's because

(ab)² = a²b².

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

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

 

[camel-man]

Ok so let me know if I am the right track if I have (9y+1)sqrt 82
i just square 9y+1 and put it under the square root with 82 and then times them both together?

 

[SteveL27]

Ok so let me know if I am the right track if I have (9y+1)sqrt 82
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 \sqrt{x^2} = x. That's because the meaning of the square root symbol is the positive number that squares to what's under the radical. So if you start with x = -5, you'll end up introducing an error.

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

 

[camel-man]

Ahh I see well I am finding the area of a surface and I need to distribute this expression into the square root due to the formula I was given

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

that is the forumula that I have to use

 

[HallsofIvy]

I think you mean "f(x)sqrt(1+ f(x) prime^2)". Please use parentheses!

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

Yes, you can write that as
\int_a^b \sqrt{f^2(x)(1+ f'^2(x))}dx

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