Discriminants and quadratics algebra problem

Question

Let $D_1$ be the discriminant of $a(x+2)^2+b(x+2)+c=0$. How does it compare to $D_2$, the discriminant of $ax^2+bx+c=0$?

Answer 1

This definition of the discriminant of a polynomial depends only on the leading coefficient and the pairwise differences of the roots. It is therefore invariant under translation, i.e. $p(x)$ and $p(x+c)$ always have the same discriminant.

Answer 2

Hint: The discriminant of the quadratic equation $$a(x+2)^2+b(x+2)+c=0$$ is given by $$D=b^2-4ac$$ the same as for $$ax^2+bx+c=0$$

Answer 3

The discriminant is essentially a value indicating how far the zeroes of a quadratic lie from the line of symmetry (do you see why?). The actual distance from the line of symmetry to the zeroes is $\frac1{2a}\sqrt{D}$. However, since one of the quadratics is just a horizontally translated version of the other, the distance to the zeroes is the same; since also their leading coefficient is the same, both $a$, we know $D_1=D_2$.

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Alternatively, we can just compute the first quadratic in terms of $x$:

$$a(x+2)^2+b(x+2)+c=ax^2 + (4a + b)x + 4a + 2b + c =0$$

Hence, the determinant is $$D_1 = (4a + b)^2 - 4a(4a + 2b + c)=b^2-4ac$$ and as it happens, $D_2=b^2-4ac$ so that $D_1=D_2$.