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platinumcaps

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How do you solve 4X^3=2.1(10^-6)?

Please show the steps by hand since we're not allowed to use a calculator. I guess my brain is ******ed after counting on a calculator too much. lol

thanks!
 
platinumcaps said:
How do you solve 4X^3=2.1(10^-6)?

Please show the steps by hand since we're not allowed to use a calculator. I guess my brain is ******ed after counting on a calculator too much. lol

thanks!

Solution #1:

4X^3 = 2.1(10^10^-6)

multiply both sides by 2
2 * (4X^3) = 2 * (2.1(10^10^-6))

8x^3 = 4.2 * 10^-6

cubic root (cbrt):
2x = (cbrt 4.2) 10^-2

We know:
(a + b)^3 = a^3 + [3 * (a^2) * b] + [3 * a * (b^2)] + b^3

(1 + a)^3 = 1 + 3a + 3a^2 + a^3
when a = 1/2:
1 + 3/2 + 3/4 + 1/8 < 4

when a = 2/3:
1 + 2 + 4/3 + 8/27 > 4.5

we want cbrt of 4.2
so number a is approximately the avg of (1/2 + 2/3) (i.e. a = (1/2 + 2/3)/ 2)
So, a = (0.5 + 0.75)/2 = 0.625

(1 + a)^3 = (1 + 0.625)^3 = (1.625)^3 = approximately 4.2

2x = (cbrt 4.2) 10^-2
2x = (approx.) (cbrt (1.625)^3) * 10^-2
x = (1.625/2) * (10^-2)

x = (approx.) 0.8 * 10^-2 = 8 * 10^-3

Solution #2:

4X^3 = 2.1 * (10^-6)?
x^3 = (2.1)/4 * 10^-6 = 525 * 10^-9
x = (cbrt 525) * 10^-3

Use a table of logarithms or successive approximations to find (cbrt 525).
 
thanks again. I really appreciate your help. I can follow your steps but I'm not sure why you decided to use (1 + a)^3 = 1 + 3a + 3a^2 + a^3 instead of another variable like (2+a)^3? Is it arbitrary?
 
platinumcaps said:
thanks again. I really appreciate your help. I can follow your steps but I'm not sure why you decided to use (1 + a)^3 = 1 + 3a + 3a^2 + a^3 instead of another variable like (2+a)^3? Is it arbitrary?

a is the decimal.

cubic root(4.2) can't be 2.a because 2 * 2 = 4 & 2 * 2 * 2 = 8
cubic root(4.2) is between 1.0 and 2.0 (i.e. 1.0 < your number <2.0 or 1.0 < 1.a < 2.0) so I used (1.0 + decimal)^3 or (1 + a)^3
 
platinumcaps said:
thanks once last time! I hope I don't have to do this much math on the dat

Use "Solution #2" for the DAT. "Solution #1" is for when you don't have a calculator and you need to find all the decimals. 🙂
 
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