Thread: x^3 + (a-1)x^2 + (9-a)x - 9 = 0

For what values of a there are exactly two real numbers that satify the equation:

x^3 + (a-1)x^2 + (9-a)x - 9 = 0

Find all possible answers.

Firstly this is what I have done.

I expand and group which leaves me with:

(x^3 - x^2) + a(x^2 - x) + 9(x - 1) = 0

Then I divide by (x-1) to get x^2 + ax + 9 = 0

Then from there I am lost, would love to know how to finish it off. Thanks!

I thought this same question was asked and answered already before?

Are you not satisfied with the answer?

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Okay, let me pitch in.

You were okay until the
"Then I divide by (x-1) to get x^2 + ax + 9 = 0"

What you did is you factored
(x^3 - x^2) + a(x^2 - x) + 9(x - 1) = 0
into
(x-1)(x^2 +ax +9) = 0

What that means now is that (x-1) is a factor of the original
x^3 + (a-1)x^2 + (9-a)x - 9 = 0 ----(1)
So,
x-1 = 0
x = 1 <----this x-value makes (1) true.
Let us check, substitute that into (1),
1^3 +(a-1)(1^2) +(9-a)(1) -9 = 0
1 +(a-1) +(9-a) -9 = 0
1 +a -1 +9 -a -9 = 0
0 = 0
See, it is true.

Then, the other factor is
x^2 +ax +9 = 0 ---------(2)
We know x=1 is okay, so, plugging it in,
1^2 +a*1 +9 = 0
1 +a +9 = 0
a = -10 --------answer?

We check that.
Plug that into (2),
x^2 +ax +9 = 0 ---------(2)
x^2 -10x +9 = 0
(x-9)(x-1) = 0
x = 9 or 1 --------two real numbers that satisfy (2).

Will those satisfy (1) also?
Should be.

Check,
>>>When x=9, and a = -10,
x^3 + (a-1)x^2 + (9-a)x - 9 = 0 ----(1)
9^3 +(-10 -1)(9^2) +(9 +10)(9) -9 =? 0
729 -11(81) +19(9) -9 =? 0
0 =? 0
Yes, so, OK.

>>>when x=1, and a = -10,
x^3 + (a-1)x^2 + (9-a)x - 9 = 0 ----(1)
1^3 +(-10 -1)(1^2) +(9 +10)(1) -9 =? 0
1 -11(1) +19(1) -9 =? 0
0 =? 0
Yes, so, OK.

Therefore, a = -10 --------answer.

Ah yes, i wasnt quite sure about the way to solve it so yes it is a repeat question. Thanks a lot ticbol.

Originally Posted by cornoth

Firstly this is what I have done.

I expand and group which leaves me with:

(x^3 - x^2) + a(x^2 - x) + 9(x - 1) = 0

Hello,

you factorized your equation. With the first factor you got the first solution, which is independent from a. The next solution comes from the quadratic factor. And there are two possible ways to solve:
1. Your quadratic equation has also the first real solution x = 1 and then you can calculate an a with which you get the other real solution of your equation as ticbol has shown.
2. It is possible to find a parameter a so that your quadratic equation has only one real solution:



*****

You'll get only one solution if the determinant equals zero:



If you put a=6 into the euation ***** you'll get the additional solution x=-3
and for a=-6 you'll get the solution x= 3

Bye