Unique Twist of quadratic equation 3

Rearranging the equation gives

$x^3-ax^2+23x-b=0$ .

We know that this equation has exactly two distinct real solutions. If we call them $m$ and $n$ , the equation above is equivalent to

$(x-m )^2 (x-n) = 0$ ,

which can be written:

$x^3-(2m+n)x^2+(m^2+2mn)x-m^2n=0$ .

By comparing the coefficients in the equations above we obtain

$a= 2m+n$

$23= m^2+2mn$

$b= m^2n$ .

We are given that $m+n=12$ , so together with $m^2+2mn-23=0$ we now have two equations with two unknowns. The quadratic equation involved gives us two pairs of solutions:

$m= 23$ , $n=-11$

and

$m=1$ , $n=11$ .

However, because the solution $m=1$ implies division by zero in the original equation, this solution must be dismissed.

By inserting $m= 23$ and $n=-11$ we get:

$a= 2m+n = 35$

$b= m^2n = -5819$ .

Thus, we have $a-b = \boxed{5854}$ .

Let $a = 11$ and $b = -35$ . Then the given equation becomes $\frac{11x^2 - 24x + 35}{x^2 - 1} = x.$ Then $\frac{(x + 1)(11x - 35)}{(x + 1)(x - 1)} = x,$ so $\frac{11x - 35}{x - 1} = x.$ Then $11x - 35 = x(x - 1) \Rightarrow x^2 - 12x + 35 = 0$ . The solutions are 5 and 7, so $a = 11$ and $b = -35$ also work.

With $x^3 - a x^2 + 23 x - b = 0,$

a = $2\alpha + \gamma$

23 = $\alpha^2 + 2 \alpha \gamma$

b = $\alpha^2 \gamma$

12 = $\alpha + \gamma$

$\alpha^2 - 24 \alpha + 23 = 0$

$\alpha = 1$ and $\gamma = 11$ OR $\alpha = 23$ and $\gamma = -11$

Original equation takes only $\alpha = 23$ and $\gamma = -11$ for $x = \alpha \neq 1$

a = $46 - 11$ = 35

b = $23^2 \times (-11)$ = -5819

a - b = 35 + 5819 = 5854

answer: $\boxed{5854}$