# Math Help - Values of a

1. ## 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 ?

2. x = { $1$, $\frac{1}{2}\,{\frac {-a+\sqrt {{a}^{2}-36\,a}}{a}}$, $\frac{1}{2}\,{\frac {-a-\sqrt {{a}^{2}-36\,a}}{a}}$}

So, 1 is a real number. Now pick an 'a' such that the 2nd and 3rd element are equal.

$a = 36$

Note: I only checked the 'a' in the real. There might be other ways of satisfying the question.

3. ## oops

Sorry but there's a typo, It should read

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

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

Thanks!

4. $x = \{ 1 , -\frac{1}{2}\,a+\frac{1}{2}\,\sqrt {{a}^{2}-36} , -\frac{1}{2}\,a-\frac{1}{2}\,\sqrt {{a}^{2}-36} \}$

Make the 2nd and 3rd element equal:

$-\frac{1}{2}\,a+\frac{1}{2}\,\sqrt {{a}^{2}-36} = -\frac{1}{2}\,a-\frac{1}{2}\,\sqrt {{a}^{2}-36}$

$a = \{6, -6\}$

5. Thanks a lot paultwang!