Simplifying an algebra question

[evosy1978]

Hello,

Ive was reading my literature today and it showed me some algebra, which i did not understand how it went from one step to the next...

What was done to simplify
V= x^2 ( 75-x^2 / 2x ) into V= x/2 ( 75-x^2 )

& then what was done to simplify ^^^

V= x/2 ( 75-x^2 ) into V= 1/2 ( 75x-x^3 )

(for anyone interested the above has now been differentiated)

I would now like to know when finding x how they go from
1/2 ( 75-3x^2 ) = 0 to this 75 = 3x^2



Thanks for any help. I know I'm missing something simple here?

 

[HallsofIvy]

Hello,

Ive was reading my literature today and it showed me some algebra, which i did not understand how it went from one step to the next...

What was done to simplify
V= x^2 ( 75-x^2 / 2x ) into V= x/2 ( 75-x^2 )

You mean V= x^2((75- x^2)/(2x)). That is, the entire "75- x^2" is divided by 2x. You can factor 1/(2x) out to get (x^2/2x)(75- x^2)= (1/2)(x)(75- x^2).

& then what was done to simplify ^^^

V= x/2 ( 75-x^2 ) into V= 1/2 ( 75x-x^3 )

Now multiply the "x" back into the 75- x^2: x/2= (1/2)(x)(75- x^2)= (1/2)(75(x)- x^2(x))= (1/2)(75x- x^3).

(for anyone interested the above has now been differentiated)

I would now like to know when finding x how they go from
1/2 ( 75-3x^2 ) = 0 to this 75 = 3x^2

Multiply both sides by 2 to get 75- 3x^2= 0
then add 3x^2 to both sides: 75= 3x^2

Thanks for any help. I know I'm missing something simple here?

Although you don't mention it, an obvious next thing to do is to divide both sides by 3:
25= x^2. What you do now depends upon what the question is! If it is to solve for x, take the square root of both sides.

 

[evosy1978]

Ok, thanks for taking the time to reply, just to let you know it can sometimes take time for things to "click in my head" lol.. I am still confused, so can I break it down and show you where I am at...

That is, the entire "75- x^2" is divided by 2x. You can factor 1/(2x) out to get (x^2/2x)(75- x^2)= (1/2)(x)(75- x^2).

I understand the last bit as you have canceled out one of the x's from the top and bottom of the fraction.
Im confused as to where the part in red 1/(2x) came from?
I also don't understand why the 2x disappears from under the 75-x^2 and then appears under the x^2? Thanks

 

[MrWarlock616]

Im confused as to where the part in red 1/(2x) came from?
I also don't understand why the 2x disappears from under the 75-x^2 and then appears under the x^2?

It's like this:

##V= x^2 (\frac{75-x^2}{2x})=\frac{1}{2x}x^2(75-x^2)=\frac{x}{2}(75-x^2)##

 

[I like Serena]

Here's a couple of the rules for working with fractions and multiplications:
$$\frac 2 3 = 2 \cdot \frac 1 3$$
$$2 \cdot 3 \cdot 4 = 4 \cdot 2 \cdot 3$$
$$\frac {3 \cdot 2} {5 \cdot 2} = \frac 3 5$$
These are the rules MrWarlock applied.