Since [tex]\alpha[/tex] and [tex]\beta[/tex] are roots of [tex]ax^2+bx+c[/tex], we can factorize the quadratic in terms of the roots as
[tex]ax^2+bx+c = a(x-\alpha)(x-\beta)[/tex]
Expanding the right side gives
[tex]ax^2+bx+c = ax^2-a(\alpha+\beta)x+a\alpha\beta[/tex]
so that
[tex]\alpha + \beta = -\dfrac ba \\\\ \alpha\beta = \dfrac ca[/tex]
A polynomial with roots [tex]\alpha^2[/tex] and [tex]\beta^2[/tex] would be
[tex](x-\alpha^2)(x-\beta^2)[/tex]
and expanding this gives
[tex]x^2 - (\alpha^2+\beta^2)x+\alpha^2\beta^2[/tex]
Now,
[tex]\alpha\beta = \dfrac ca \implies \alpha^2\beta^2 = (\alpha\beta)^2 = \dfrac{c^2}{a^2}[/tex]
and
[tex](\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \\\\ \implies \alpha^2+\beta^2 = \left(-\dfrac ba\right)^2 -2\left(\dfrac ca\right) = \dfrac{b^2-2ac}{a^2}[/tex]
So we can write the second quadratic in terms of [tex]a,b,c[/tex] as
[tex](x-\alpha^2)(x-\beta^2) = x^2 - \dfrac{b^2-2ac}{a^2}x + \dfrac{c^2}{a^2}[/tex]
and to make things look cleaner, scale the whole expression by [tex]a^2[/tex] to get
[tex]\boxed{a^2x^2 + (2ac-b^2)x + c^2}[/tex]
