a³ + 3a = 140 (1 Viewer)

[IAU001]

I was just looking over my old yearly maths test paper when i saw this question. school's out so i can't ask my teacher. I would really appreciate it if you could help me on how to solve it. i'm kinda slow so a step by step procedure would be REALLY appreciated

thanks in advance

 

[Affinity]

well the way you are meant to do it:
you want to sovle a^3 + 3a - 140 = 0
after a bit of trying you discover that a=5 is a solution.
then you factorize it:
(a-5)(a^2 + 5a + 28) = 0
and since a^2 + 5a + 28 = 0 has no real valued solutions, a = 5 is the only real solution.

the general way:
let a = (x+y)
then you get:

(x^3 + y^3) + (3xy+3)(x+y) = 140
so, if we can get x^3 + y^3 = 140 and xy = -1, we has a solution.

so x=-1/y
y^3 - 1/y^3 = 140
(y^3)^2 - 140(y^3) - 1 = 0
y^3 = {140 [+/-] sqrt(140^2 + 4)}/2
y = cuberoot({140 [+/-] sqrt(140^2 + 4)}/2) etc.

 

[jyu]

How do you derive these formulae? Is it a good idea for sec students to learn the derivations?

 

[bos1234]

that cubic forumla method is confusing...... i think ill stick to the remainder theorme..

 

[Riviet]

that cubic forumla method is confusing...... i think ill stick to the remainder theorme..

Don't worry about it if it's confusing you, I never used it and it's not in the syllabus.

 

[IAU001]

thanks for that