JexteLox

Advanced Algebra


if *x* can be expressed as y^2, where y is a positive integer, then let *x*= y^3 / 2. For example, since 9= 3^2, *9*= 3^3 /2 = 27/2 = 13.5.

if *m*= 4, what is the value of *4m*?
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very lost here. explain step by step ?

JexteLox

Math geniuses, now is the time

Baelor

Okay, here's the problem. You have y twice, in two places that don't work together (I was looking at this last night).

To say that *x* can be expressed as y^2 implies that *x*=y^2, but also equals y^3/2. That doesn't work because the example provided doesn't satisfy both of those equalities. How about this?

If x can be expressed as y^2, then *x*=y^3/2. This should make the question more reasonable.

*m*=4
y^3/2=4 for some y
y=2 by solving.

So m=y^2=4, which means *4m*=*16*

16=4^2, so *4m* = *16* = 4^3/2 = 64/2 = 32

Is that correct?

arnoc

x can be expressed as y^2; thus, sqr(x) = +/- y, an integer; from this, x is a perfect square
*x* = y^3 / 2

Since x is simply a variable, you can replace it with any number or any other symbol

*m* = 4
*m* = y^3 / 2 -> 4 = y^3 / 2 -> 8 = y^3 -> y = 8^(1/3) = 2

By definition, m is y^2, so m is also 4; so, 4m = 16

16 = y^2 -> y = 4
*16* = 4^3 / 2 = 64 / 2 = 32

*4m* = 32

gcf101

Another way:

*y^2* = y^3 /2 <------- (A)

I.
*m* = 4 means
m=y^2 and y^3 /2 = 4

II.
4m = (4)y^2 = (2y)^2
*4m* = *(2y)^2*
*(2y)^2* = (2y)^3 /2
*(2y)^2* = (8)y^3 /2
*(2y)^2* = (8)4
*(2y)^2* = 32
*4m* = 32.
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To make part II more clear (but longer):
4m = (4)y^2 = (2y)^2
Let n=2y
4m = n^2 <----------- (B)
*4m* = *n^2*
*n^2* = n^3 /2 <------------ from (A)
*n^2* = (2y)^3 /2
*n^2* = (8)y^3 /2
*n^2* = (8)4
*n^2* = 32
*4m* = 32.

gluttony

oh shoot,i wanted to answer,but already enough explanations