Troll Quadratic

Since $a,b$ are roots of the equation, then we have: $\left\{ \begin{array}{lr} (1) : f(a) = 2a^2 + b = 0 \\ (2) : f(b) = b^2 + ab + b = 0 \end{array} \right.$ From $(1)$ we see that $b = -2a^2$ .

From $(2)$ , we see that $b(b+a+1) = 0$ . Substituting the above equation gives us $-2a^2(-2a^2+a+1) = 0$ This equation has three distinct solutions: $a = 0, 1, -\frac{1}{2}$ . This gives us three distinct solutions: $(0,0), (1,-2), \left(-\dfrac{1}{2}, -\dfrac{1}{2}\right)$ .

Hence there are a total of $\boxed{3}$ solutions.

We proceed as expected: obviously, we have $x^2+ax+b=(x-a)(x-b)$

Expanding gives $x^2+ax+b=x^2-(a+b)x+ab$ which means $a=-(a+b)$ and $b=ab$ .

If $b=0$ , then $a=0$ and we have our first solution $(a,b)=(0,0)$ .

If $b\ne 0$ , then dividing by $b$ in the second equation gives $a=1$ which gives $b=-2$ so our second solution is $(a,b)=(1,-2)$ .

This is where the tricky part comes in: I never said that $a$ and $b$ were the two roots of the quadratic, I just said that they were roots. Therefore, we have one more case where $a$ and $b$ are referring to the same root .

Thus, in this case $a=b$ and we have the equation $x^2+ax+a=(x-a)(x-c)$ for some $c$ .

Expanding gives $x^2+ax+a=x^2-(a+c)x+ac$ which implies $a=-(a+c)$ and $a=ac$ .

If $a=0$ , then $c=0$ and we get a solution we already found.

If $a\ne 0$ , then dividing both sides of the second equation by $a$ gives $c=1$ so $a=-\dfrac{1}{2}$ which gives the solution $(a,b)=\left(-\dfrac{1}{2},-\dfrac{1}{2}\right)$

This gives a total of $\boxed{3}$ ordered pairs of solutions.

Because $a$ and $b$ are roots of the equation $f(x) = x^2 + ax + b = 0$ , we have: $f(a) = 2a^2 + b = 0 \\ f(b) = b^2 + ab + b = 0$

The first equation can be rewritten as:

$b = -2a^2$

Substituting it into the second equation:

$4a^4 - 2a^3 - 2a^2 = 0$

$\Leftrightarrow a^2(4a^2 - 2a - 2) = 0$

$\Leftrightarrow a^2(a - 1)(a+\frac {1} {2}) = 0$

$\Leftrightarrow a = 0$ or $a = 1$ or $a =- \frac {1} {2}$

Therefore there are three solutions of $(a,b)$

Consider equal roots case, hence $a=b$ . Then, we have $a^2+a^2+a=0\Rightarrow a=0, -\frac{1}{2}$ . Hence, two ordered pairs here, that is $(a,b)=(0,0),(-\frac{1}{2},-\frac{1}{2})$ .

Now consider different roots case. Since the sum of roots is $-a$ and product of roots is $b$ , we have $a+b=-a$ and $ab=b$ . Solving the system of equation gives the ordered pairs $(a,b)=(1,2),(0,0)$ .

In total, there are $\boxed{3}$ $\textbf{distinct}$ ordered pairs.

We mustn't compare half way in this question by treating $dependent$ $variables$ as independent variables. To simplify the solving process, let's just discuss why (a, b) of (0, 0), (1, -2) and also (-0.5, -0.5) are all suitable.

Both $x^2 + a x + b = 0$ and $x^2 - (a + b) x + a b = 0$ ought to be satisfied, by ignoring y as how roots are defined for a given polynomial.

Four equations as follows are all satisfied by (a, b) of values either (0, 0), (1, -2) or ( $-\frac12, -\frac12$ ):

1) $a^2 + a a + b = 0$

2) $a^2 - (a + b) a + a b = 0$

3) $b^2 + a b + b = 0$

4) $b^2 - (a + b) b + a b = 0$

Since all of the 3 ordered pairs introduced satisfy as dependent variables to equations, they are all valid cases.

Answer: $\boxed{3}$