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.

 

[buchanan]

Yeah. That's OK.

You can test the divisors of 140. Eventually you will discover that 5 works. And that's as much as you need for this question.

So what are the divisors of 140?

±1, ±2, ±4, ±5, ±7, ±10, ±14, ±20 ±28, ±35, ±70, ±140

but there's no need to test the negative ones, so try only 1,2,4,5,7,10,14,20,28,35,70,140 - but once you find one you can factorise, so starting from the smallest ones,

1³+3(1)-140=-136
2³+3(2)-140=-126
4³+3(4)-140=-64
5³+3(5)-140=0 STOP and factorise:

a³+3a-140=(a-5)(a²+5a+28)=0, as affinity said and for the quadratic factor, &Delta;=-87 < 0 so there is only 1 real root and it's a=5.

However this method doesn't always work and there is a more general formula for cubics:

and using this formula we see that for your question,

I made a thread on cubics and quartics at
http://community.boredofstudies.org...acurricular-topics/99063/cubics-quartics.html

Here is an attachment of a more complete version of affinity's method:

 

[buchanan]

Challenge: Now see if you can solve a⁴+3a=140

<a href="http://users.tpg.com.au/nanahcub/ans.gif">click here for the answer</a> using <a href="http://users.tpg.com.au/nanahcub/quartic.gif">this formula</a>

 

[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.

 

[buchanan]

There are proofs at

http://mathworld.wolfram.com/CubicFormula.html

http://mathworld.wolfram.com/QuarticEquation.html

 

[buchanan]

And here's an article which appeared on it in UNSW's Parabola Magazine in 2005:

 

[IAU001]

thanks for that