Quadratics

What you need to know

• Quadratic formula. The roots of $ax^2+bx+c=0$ are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. Example: $2x^2+3x-2=0$ gives $x=\frac{-3\pm5}{4}$, so $x=\frac12$ or $x=-2$.
• The discriminant $D=b^2-4ac$ is the part under the square root. If $D$ is positive there are two real roots; if $D=0$ there is one repeated root (a double root); if $D$ is negative there are no real roots. Example: $x^2-6x+9$ has $D=0$ and the double root $3$; $x^2+2x+5$ has $D=-16$ and no real roots.
• When $a$, $b$, $c$ are integers and $D$ is a perfect square, the roots are rational. Example: above, $D=9+16=25$, and the roots $\frac12$, $-2$ are rational.
• Vieta's Formulas. If $r$ and $s$ are the roots, then $r+s=-\dfrac ba$ and $rs=\dfrac ca$. Example: for $3x^2-12x+7=0$, $r+s=4$ and $rs=\frac73$.
• Two identities that use only $r+s$ and $rs$: $r^2+s^2=(r+s)^2-2rs$, and $(r-s)^2=(r+s)^2-4rs$, which also equals $\dfrac{D}{a^2}$. Example: for $x^2-5x+3=0$, $r^2+s^2=25-6=19$ and $(r-s)^2=25-12=13=D$.
• Completing the square: $ax^2+bx+c=a\left(x+\dfrac b{2a}\right)^2+c-\dfrac{b^2}{4a}$. So the vertex (the top or bottom of the parabola) is at $x=-\dfrac b{2a}$, halfway between the two roots. Example: $2x^2-8x+3=2(x-2)^2-5$, so the lowest point is at $x=2$, where the value is $-5$.
• Any expression that stays the same when you swap $r$ and $s$ can be written using only $r+s$ and $rs$. Example: $\frac1r+\frac1s=\frac{r+s}{rs}$.

How AMC 10 tests it

• Problems 5–12: "sum of the roots" or "product of the roots," lightly disguised (a common factor, or an equation not yet in standard form).
• Problems 10–18: "for how many integers $k$ does $x^2+kx+n=0$ have integer roots?" By Vieta, the roots multiply to $n$, so list the factor pairs of $n$.
• Problems 12–18: an expression in the roots that treats them the same way ($r^2+s^2$, $\frac1r+\frac1s$, $r^3+s^3$), or a new quadratic whose roots are built from the old ones.
• A largest/smallest-value question from a word or geometry setup that is really the vertex of a parabola.
• "A line meets a parabola in exactly one point": set the two equal and look at 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 swapping the roots doesn't change what is asked, use Vieta's Formulas and never solve for the roots.
3. For integer-root questions, list the factor pairs of the constant term, including negative pairs, and read off the possible sums.
4. For a maximum or minimum, complete the square or use $x=-\frac b{2a}$.
5. To count real solutions or crossing points, compute the discriminant and check whether it is positive, zero, or negative.

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) $\frac{248}{5}$

The roots are messy ($\frac{7\pm\sqrt{29}}{2}$), so we avoid them. By Vieta's Formulas, $r+s=7$ and $rs=5$.

Put the two fractions over the common denominator $rs$:
$$\frac{r^2}{s}+\frac{s^2}{r}=\frac{r^3+s^3}{rs}.$$
To find $r^3+s^3$, expand $(r+s)^3=r^3+3r^2s+3rs^2+s^3=r^3+s^3+3rs(r+s)$. So
$$r^3+s^3=(r+s)^3-3rs(r+s)=343-3\cdot5\cdot7=343-105=238.$$
Dividing by $rs=5$ gives $\frac{238}{5}$. The answer is $\boxed{\textbf{(C)}\ \frac{238}{5}}$.

Pitfalls

• Using Vieta's Formulas before the equation is in standard form. Example: $2x^2=5x-3$ is $2x^2-5x+3=0$, so $b=-5$, not $5$, and the roots add to $\frac52$.
• Dropping the minus sign in $r+s=-\frac ba$. Example: the roots of $x^2+5x+6=0$ are $-2$ and $-3$, which add to $-5$.
• Forgetting to divide by the leading coefficient in both the sum and the product. Example: the roots of $2x^2-5x+3=0$ multiply to $\frac32$, not $3$.
• Mixing up "two real roots" and "two different real roots." "Two different (distinct) real roots" always means $D>0$. Plain "real roots" usually allows the two roots to be equal, which means $D\ge0$. Read the wording carefully.

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)