# Values of a

• May 23rd 2005, 02:22 AM
cornoth
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 ?
• May 23rd 2005, 08:34 AM
paultwang
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.
• May 24th 2005, 03:44 AM
cornoth
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!
• May 24th 2005, 10:50 AM
paultwang
$x = \{ 1 , -\frac{1}{2}\,a+\frac{1}{2}\,\sqrt {{a}^{2}-36} , -\frac{1}{2}\,a-\frac{1}{2}\,\sqrt {{a}^{2}-36} \}$

Make the 2nd and 3rd element equal:

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

$a = \{6, -6\}$
• Jun 19th 2005, 07:48 AM
cornoth
Thanks a lot paultwang! :o