Quadratics

What you need to know

• Quadratic formula: the roots of $ax^2+bx+c=0$ are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. The discriminant $D=b^2-4ac$ is positive, zero, or negative according as there are two real roots, one double root, or no real roots; with integer coefficients, $D$ a perfect square means rational roots.
• Vieta: if $r,s$ are the roots, then $r+s=-\dfrac ba$ and $rs=\dfrac ca$. Hence $r^2+s^2=(r+s)^2-2rs$ and $(r-s)^2=(r+s)^2-4rs=\dfrac{D}{a^2}$.
• Completing the square: $ax^2+bx+c=a\left(x+\dfrac b{2a}\right)^2+c-\dfrac{b^2}{4a}$; the vertex is at $x=-\dfrac b{2a}$, midway between the roots.
• Any expression symmetric in $r$ and $s$ can be written using only $r+s$ and $rs$.

How AMC 10 tests it

• Problems 5–12: "sum of the roots" or "product of the roots" after a light disguise (a common factor, an equation not yet in standard form).
• Problems 10–18: "for how many integers $k$ does $x^2+kx+n=0$ have integer roots?" — a factor-pair search on $n$ via Vieta.
• Problems 12–18: a symmetric function of the roots ($r^2+s^2$, $\frac1r+\frac1s$, $r^3+s^3$), or a new quadratic whose roots are built from the old ones.
• A max/min question from a word or geometry setup that is really the vertex of a parabola.
• "Line meets parabola in one point": set equal and analyze the discriminant.

Standard approaches

1. Put the equation in the form $ax^2+bx+c=0$ (clear fractions, move everything to one side) before using any formula.
2. If the question is symmetric in the roots, use Vieta and never solve for the roots.
3. For integer-root questions, list factor pairs of the constant term, including negatives, and read off the possible sums.
4. For a maximum or minimum, complete the square or use $x=-\frac b{2a}$.
5. For counting real solutions or intersections, compute the discriminant and study its sign.

Worked example

Let $r$ and $s$ be the roots of $x^2-7x+5=0$. What is $\dfrac{r^2}{s}+\dfrac{s^2}{r}$?

(A) $\frac{218}{5}$ (B) $\frac{228}{5}$ (C) $\frac{238}{5}$ (D) $49$ (E) $\frac{248}{5}$

By Vieta, $r+s=7$ and $rs=5$. Combine over a common denominator:
$$\frac{r^2}{s}+\frac{s^2}{r}=\frac{r^3+s^3}{rs}=\frac{(r+s)^3-3rs(r+s)}{rs}=\frac{343-105}{5}=\frac{238}{5}.$$
The answer is $\boxed{\textbf{(C)}\ \frac{238}{5}}$.

Pitfalls

• Applying Vieta before the equation is in standard form (reading $2x^2=5x-3$ as $b=5$).
• Dropping the minus sign in $r+s=-\frac ba$.
• Forgetting that the leading coefficient divides both the sum and the product.
• Miscounting a double root: "two real roots" and "distinct real roots" are different conditions ($D\ge0$ versus $D>0$).

Traps that recur

• Maximizing the sum of the four solutions of p(p(x)) = 0, or evaluating p at 1 for a polynomial found from a wrong tangency root (using the larger root as the vertex value). (2021 AMC Fall 10A #25)
• Using only the positive factor pairs of 4, which yields a = 9 and 8 and the sum 17 (choice D). (2015 AMC 10A #23)
• Finding the ratio a/c = 7 + 4sqrt3 and reporting it (or its negative) instead of the root -b/(2a). (2013 AMC 10B #19)
• Dividing the sum of reciprocals by 2025 (the number of factors) rather than 4050 (the number of roots), which turns the answer into -3/4 and tempts a guess at the nearest choice. (2025 AMC 10A #18)