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- Mar 30th 2011, 06:06 PM #1

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## A^3 + a + - 10 = 0

My directions were:

Solve algebraically

A^3 - B = 5

A + B = 5

I added the equations and got A^3 + A = 10

I realize 2 is the solution but cannot figure out how to do it algebraically. I don't see a way to factor by grouping and my teacher never showed us the general cubic formula.

- Mar 30th 2011, 06:09 PM #2

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- Mar 30th 2011, 06:43 PM #3
I'm not sure why your teacher would give you this one if you don't know how to approach it. However there is the rational root theorem. However if you really need an algebraic solution, there is always Cardano's method. (Go to the subheading "Cardano's method.") However this is pretty rough work for the High School level, so I'm pretty sure that you will want to give this one a miss.

-Dan

- Mar 30th 2011, 06:55 PM #4

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Sudharaka, if I take your suggestion I get (A+2)(A^2 + 2A + 5) = 0 but the problem is I knew to divide by a+2 only by inspection. Thus, that's not following the directions of solve algebraically.

I think either she didn't want me to use purely analytic methods once I got to a^3 + a = 10 or it might be a typo actually where it was supposed to be A^1 or A^2 instead of A^3. Or maybe by solve algebraically she just wanted me to get to the A^3 + A = 10 part and then maybe she wanted me to do that by inspection.

- Apr 1st 2011, 04:41 AM #5

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- Apr 1st 2011, 06:58 AM #6

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But I can see no reason for saying that seeing that x= 2 is a solution "by inspection" is NOT an algebraic method!

(NOT "(A+ 2)").

But if you really feel you need to show "factoring by grouping", you coud write

.

(Actually, I got that from

, and but you don't need to show that to your teacher!)

- Apr 1st 2011, 09:44 AM #7

- Apr 1st 2011, 10:37 AM #8

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- Apr 1st 2011, 10:45 AM #9

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Descartes rule of signs tells us that this has one positive root and zero negative roots.

The rational roots theorem tells us that if this has a rational root it is one of 1,2,5 (need only the positive candidates here because we know that there is only one real root and it is positive).

Checking tells us that A=2 is a root, and it is the only real root. The complex roots can be found by taking out the factor (A-2) which leaves a quadratic that can be solved using the quadratic formula.

Back substitution into the original equations shows that A=2, B=3 is indeed a solution to the original equations

(this is an algebraic solution in that Descartes rule of signs, the rational roots theorem and synthetic or polynomial division are/were all included in what would be called college algebra)

CB