Values of a

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

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

I assume the second term was x^2 and not x^3.

Exactly 2 real solutions means two overlapping at a minimum or maximum, and one other. In order to have a minimum and maximum, the first derivative must have two real solutions, that means a non-negative discriminant. If my calc is correct then this is possible with a < -5.62 and a > 4.62.

By observation, x=1 is a solution of the polynomial. When you divide out (x-1), you are left with:

x^2+a*x+9 = 0

Solving this equation using the quadratic formula, you get:

x = [-a +/- SQRT(a^2-36)]/2

Since we already have one root (x=1), we find the second (repeated) root by setting the argument under the square root sign equal to zero:

a^2 - 36 = 0
a = +/- 6

Neither solution is equal to 1, so we are assured we have only 2 roots. When we set a=6, the polynomial becomes:

(x-1)(x+3)^2 = 0
(x-1)(x^2+6x+9) = 0
x^3+5x^2+3x-9 = 0

And when we set a=(-6), the polynomial becomes:

(x-1)(x-3)^2 = 0
(x-1)(x^2-6x+9) = 0
x^3-7x^2+15x-9 = 0

So the solutions are a=6 or a=(-6).

Originally posted by PBE6
By observation, x=1 is a solution of the polynomial. When you divide out (x-1), you are left with:

x^2+a*x+9 = 0

Solving this equation using the quadratic formula, you get:

x = [-a +/- SQRT(a^2-36)]/2

Since we already have one root (x=1), we find the second (repeated) root by setting the argument under the square root sign equal to zero:

a^2 - 3 ...[text shortened]... (x-1)(x-3)^2 = 0
(x-1)(x^2-6x+9) = 0
x^3-7x^2+15x-9 = 0

So the solutions are a=6 or a=(-6).

Originally posted by Mephisto2
I assume the second term was x^2 and not x^3.

Exactly 2 real solutions means two overlapping at a minimum or maximum, and one other. In order to have a minimum and maximum, the first derivative must have two real solutions, that means ...[text shortened]... calc is correct then this is possible with a < -5.62 and a > 4.62.