halp with nightmare question

>>354981
two equations with two variables.

solve the second equation for a (a = ...) and insert this in the first. You get a quadratic equation that only depends on b. You can solve for b.

Or you can see that 3.75 = 15/4 and go from there.

>>354996
I already tried the quadratic equation thing, senpai. The bhaskara's delta is not exact and even if i try to use the calculator the result does not match with any item

>>354981
Could you redraw this so the numbers/letters are clear? I can't even tell what you're trying to do.

File: WP_20170729_20_21_45_Pro.jpg (1MB, 3264x1840px) Image search: [Google]

1MB, 3264x1840px

>>355075
Here you go senpai

>>355075
Sorry for taking long, connections shit here

>>355109
>>355111
That's better, though your a and 2 still look way too similar. It's also weird because the first box seems totally irrelevant, and then you name the next cubes "2" and "3" and put the numbers right on top of the cubes. Anyway,

a*b = 3.75
a-b = 1, therefore b=a-1.
a(a-1) = 3.75
a^2 - a - 3.75 = 0
Quadratic formula says:
a = (-(-1) +- sqrt(1 - 4(1)(-3.75))/2*1
a = 1 +- sqrt(1+15)/2
a = .5 +- 2
And since these are both real side lengths, we only take the positive answer.
a = 2.5
b = 3.75/2.5 = 1.5
a^3 - b^3 = 12.25

I'm not sure what your approach was, but this is in no way a "nightmare question" and it sounds like you massively overcomplicated this. I guess you tried the simple thing, made a minor error and didn't catch it, and decided the entire solution method was invalid. Happens to all of us occasionally.

>>355109
as I said, you can solve this with a quadratic equation.
a = b + 1
ab = (b+1)b = 3.75 = 15/4
b^2 + b + 1/4 - 1/4 = 15/4
(b + 0.5)^2 - 1/4 = 15/4
(b + 0.5)^2 = 16/4 = 4
b + 0.5 = 2
b = 1.5 = 3/2

or you can simply see that 3.75 = 15/4 = (3/2)(5/2) and the difference between 5/2 and 3/2 is 1 as required.

>>354981 (OP)
here's an alternate solution

a^3 - b^3 is a difference of cubes
a^3 - b^3 = (a - b)(a^2 + ab + b^2)

you're given (a - b), but what about (a^2 + ab + b^2)?

(a - b)^2
= a^2 - 2ab + b^2 = 1m^2
>now add 3ab to this
a^2 - 2ab + b^2 + 3ab = 1m^2 + 3*3.75m^2
= a^2 + ab + b^2 = 12.25m^2

a^3 - b^3
= (a - b)(a^2 + ab + b^2)
= (1m)(12.25m^2)
= 12.25m^3

>>355122
>>355129
I don't see how either of these are better than just using the quadratic formuoli.

>>355121
Seeing like this i guess it was really a simple mistake along the calculations !!!
Oh well, algebra has these things, i guess.

>>355121
About the useless box, i think it was just a false flag played by the problem. I just copied it.