Math problem (cubic equation-related)

**Altovirator** · January 1st 2010, 04:55 AM

Find the exact value of x so that the equation

has exactly 2 solutions when solved for a.

**HallsofIvy** · January 1st 2010, 05:02 AM

Originally Posted by Altovirator

Find the exact value of x so that the equation

has exactly 2 solutions when solved for a.

You mean, I take it, 2 distinct real solutions. In that case, the equation must be of the form $(a-u)^2(a-v)= a^3- a- x= 0$ . for some u and v.

Multiplying that out, we get $a^3- (2u+v)a^2+ (u^2+ 2uv)a- u^2v= a^3- a- x= 0$ . We must have 2u+v= 0, $u^2+ 2uv= -1$ and $u^2v= x$ . Solve the first two equations for u and v and then get x from the third equation.

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