Write the number of quadratic equations, with real roots,

Let $\alpha$ and $\beta$ be the real roots of the quadratic equation $a x^{2}+b x+c=0$.

On squaring these roots, we get:

$\alpha=\alpha^{2} \quad$ and $\quad \beta=\beta^{2}$

$\Rightarrow \alpha(1-\alpha)=0$ and $\beta(1-\beta)=0$

$\Rightarrow \alpha=0, \alpha=1$ and $\beta=0,1$

Three cases arise: 

(i) $\alpha=0, \beta=0$

(ii) $\alpha=1, \beta=0$

(iii) $\alpha=1, \beta=1$

$(i) \alpha=0, \beta=0$

$\therefore \alpha+\beta=0$ and $\alpha \beta=0$

So, the corresponding quadratic equation is,

$x^{2}-(\alpha+\beta) x+\alpha \beta=0$

$\Rightarrow x^{2}=0$

$(i i) \alpha=0, \beta=1$

$\alpha+\beta=1$

$\alpha \beta=0$

So, the corresponding quadratic equation is,

$x^{2}-(\alpha+\beta) x+\alpha \beta=0$

$\Rightarrow x^{2}-x+0=0$

$\Rightarrow x^{2}-x=0$

$(i i i) \alpha=1, \beta=1$

$\alpha+\beta=2$

$\alpha \beta=1$

So, the corresponding quadratic equation is,

$x^{2}-(\alpha+\beta) x+\alpha \beta=0$

$\Rightarrow x^{2}-2 x+1=0$

Hence, we can construct 3 quadratic equations.