Question

please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve it please solve​

Answer 1

Since $\alpha$ and $\beta$ are roots of $ax^2+bx+c$, we can factorize the quadratic in terms of the roots as

$ax^2+bx+c = a(x-\alpha)(x-\beta)$

Expanding the right side gives

$ax^2+bx+c = ax^2-a(\alpha+\beta)x+a\alpha\beta$

so that

$\alpha + \beta = -\dfrac ba \\\\ \alpha\beta = \dfrac ca$

A polynomial with roots $\alpha^2$ and $\beta^2$ would be

$(x-\alpha^2)(x-\beta^2)$

and expanding this gives

$x^2 - (\alpha^2+\beta^2)x+\alpha^2\beta^2$

Now,

$\alpha\beta = \dfrac ca \implies \alpha^2\beta^2 = (\alpha\beta)^2 = \dfrac{c^2}{a^2}$

and

$(\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \\\\ \implies \alpha^2+\beta^2 = \left(-\dfrac ba\right)^2 -2\left(\dfrac ca\right) = \dfrac{b^2-2ac}{a^2}$

So we can write the second quadratic in terms of $a,b,c$ as

$(x-\alpha^2)(x-\beta^2) = x^2 - \dfrac{b^2-2ac}{a^2}x + \dfrac{c^2}{a^2}$

and to make things look cleaner, scale the whole expression by $a^2$ to get

$\boxed{a^2x^2 + (2ac-b^2)x + c^2}$