AlgebraDifficulty 25-75640 tagged problems

Vieta's Formulas

Vieta's Formulas is the habit of connecting polynomial roots to coefficient sums and products to avoid solving explicitly. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\text{For } ax^2 + bx + c = 0 \text{ with roots } r, s:\quad r + s = -\frac{b}{a}, \qquad rs = \frac{c}{a}.$$

A polynomial's coefficients encode the sum and product of its roots directly, so you can use root relationships without ever solving for the roots.

Why It Works

1
If $r$ and $s$ are the roots of $ax^2 + bx + c = 0$, then $ax^2 + bx + c = a(x - r)(x - s)$.
2
Expand the right side: $a(x^2 - (r+s)x + rs) = ax^2 - a(r+s)x + a\,rs$.
3
Match coefficients with $ax^2 + bx + c$: the $x$ term gives $b = -a(r+s)$, and the constant gives $c = a\,rs$.
4
Divide by $a$: $r + s = -\frac{b}{a}$ and $rs = \frac{c}{a}$. The same coefficient-matching extends to higher degrees (alternating signs on elementary symmetric sums).

Worked Examples

Example 1

The roots of $x^2 - 7x + 10 = 0$ are $r$ and $s$. Find $r^2 + s^2$ without solving for the roots.

1
By Vieta, $r + s = 7$ and $rs = 10$.
2
Use the identity $r^2 + s^2 = (r + s)^2 - 2rs$.
3
Substitute: $r^2 + s^2 = 7^2 - 2\cdot 10 = 49 - 20 = 29$.

Answer:

$r^2 + s^2 = 29$.

Warm-up

The roots of $2x^2 - 3x + 5 = 0$ are $r$ and $s$. Find $\dfrac{1}{r} + \dfrac{1}{s}$.

1
By Vieta, $r + s = -\dfrac{-3}{2} = \dfrac{3}{2}$ and $rs = \dfrac{5}{2}$.
2
Combine the fractions: $\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs}$.
3
Substitute: $\dfrac{3/2}{5/2} = \dfrac{3}{5}$.

Answer:

$\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{3}{5}$.

Contest level

Let $r, s, t$ be the roots of $x^3 - 6x^2 + 11x - 6 = 0$. Compute $r^2 + s^2 + t^2$.

1
Cubic Vieta (monic): $r + s + t = 6$, $rs + st + tr = 11$, $rst = 6$.
2
Use $r^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + st + tr)$.
3
Substitute: $6^2 - 2(11) = 36 - 22 = 14$.

Answer:

$r^2 + s^2 + t^2 = 14$.

Olympiad / Challenge

Let $a, b, c$ be the roots of $x^3 - 3x^2 + 5x - 7 = 0$. Compute $a^2 b^2 + b^2 c^2 + c^2 a^2$.

1
Vieta gives $e_1 = a + b + c = 3$, $e_2 = ab + bc + ca = 5$, $e_3 = abc = 7$.
2
Use the identity $(ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c)$, i.e. $e_2^2 = \sum a^2 b^2 + 2 e_3 e_1$.
3
Solve for the target: $\sum a^2 b^2 = e_2^2 - 2 e_1 e_3 = 25 - 2(3)(7) = 25 - 42 = -17$.

Answer:

$a^2 b^2 + b^2 c^2 + c^2 a^2 = -17$.

Going Deeper

Generalization: for $a_n x^n + \cdots + a_0$ with roots $r_i$, the $k$th elementary symmetric sum is $e_k = (-1)^k \dfrac{a_{n-k}}{a_n}$ — so $\sum r_i = -\frac{a_{n-1}}{a_n}$, $\prod r_i = (-1)^n \frac{a_0}{a_n}$, and everything in between alternates sign.
Where it appears: AIME problems that ask for $\sum r_i^2$, $\sum \frac{1}{r_i}$, or a polynomial with transformed roots — solvable WITHOUT finding the (often irrational) roots. Newton's sums extend Vieta to arbitrary power sums.
Pitfall: watch the leading coefficient — the formulas have $a$ in the denominator, so $r + s = -\frac{b}{a}$, NOT $-b$, unless the polynomial is monic. And mind the alternating signs in higher degree: a dropped sign on $e_2$ or $e_3$ is the classic Vieta error.

Spot the Signal

Look for problems where the key step is connecting polynomial roots to coefficient sums and products to avoid solving explicitly.
You can describe the hard part as connecting polynomial roots to coefficient sums and products to avoid solving explicitly, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for vieta's formulas, then rewrite the givens around it.
Name the relation that makes Vieta's Formulas legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Vieta's Formulas just because the surface looks familiar; verify the required condition first.
Applying Vieta's Formulas because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let vieta's formulas reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is connecting polynomial roots to coefficient sums and products to avoid solving explicitly.

Back to techniques

AlgebraDifficulty 25-75640 tagged problems

Vieta's Formulas

Practice this technique Match my level

The Key Result

$$\text{For } ax^2 + bx + c = 0 \text{ with roots } r, s:\quad r + s = -\frac{b}{a}, \qquad rs = \frac{c}{a}.$$

A polynomial's coefficients encode the sum and product of its roots directly, so you can use root relationships without ever solving for the roots.

Why It Works

1
If $r$ and $s$ are the roots of $ax^2 + bx + c = 0$, then $ax^2 + bx + c = a(x - r)(x - s)$.
2
Expand the right side: $a(x^2 - (r+s)x + rs) = ax^2 - a(r+s)x + a\,rs$.
3
Match coefficients with $ax^2 + bx + c$: the $x$ term gives $b = -a(r+s)$, and the constant gives $c = a\,rs$.
4
Divide by $a$: $r + s = -\frac{b}{a}$ and $rs = \frac{c}{a}$. The same coefficient-matching extends to higher degrees (alternating signs on elementary symmetric sums).

Worked Examples

Example 1

The roots of $x^2 - 7x + 10 = 0$ are $r$ and $s$. Find $r^2 + s^2$ without solving for the roots.

1
By Vieta, $r + s = 7$ and $rs = 10$.
2
Use the identity $r^2 + s^2 = (r + s)^2 - 2rs$.
3
Substitute: $r^2 + s^2 = 7^2 - 2\cdot 10 = 49 - 20 = 29$.

Answer:

$r^2 + s^2 = 29$.

Warm-up

The roots of $2x^2 - 3x + 5 = 0$ are $r$ and $s$. Find $\dfrac{1}{r} + \dfrac{1}{s}$.

1
By Vieta, $r + s = -\dfrac{-3}{2} = \dfrac{3}{2}$ and $rs = \dfrac{5}{2}$.
2
Combine the fractions: $\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs}$.
3
Substitute: $\dfrac{3/2}{5/2} = \dfrac{3}{5}$.

Answer:

$\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{3}{5}$.

Contest level

Let $r, s, t$ be the roots of $x^3 - 6x^2 + 11x - 6 = 0$. Compute $r^2 + s^2 + t^2$.

1
Cubic Vieta (monic): $r + s + t = 6$, $rs + st + tr = 11$, $rst = 6$.
2
Use $r^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + st + tr)$.
3
Substitute: $6^2 - 2(11) = 36 - 22 = 14$.

Answer:

$r^2 + s^2 + t^2 = 14$.

Olympiad / Challenge

Let $a, b, c$ be the roots of $x^3 - 3x^2 + 5x - 7 = 0$. Compute $a^2 b^2 + b^2 c^2 + c^2 a^2$.

1
Vieta gives $e_1 = a + b + c = 3$, $e_2 = ab + bc + ca = 5$, $e_3 = abc = 7$.
2
Use the identity $(ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c)$, i.e. $e_2^2 = \sum a^2 b^2 + 2 e_3 e_1$.
3
Solve for the target: $\sum a^2 b^2 = e_2^2 - 2 e_1 e_3 = 25 - 2(3)(7) = 25 - 42 = -17$.

Answer:

$a^2 b^2 + b^2 c^2 + c^2 a^2 = -17$.

Going Deeper

Generalization: for $a_n x^n + \cdots + a_0$ with roots $r_i$, the $k$th elementary symmetric sum is $e_k = (-1)^k \dfrac{a_{n-k}}{a_n}$ — so $\sum r_i = -\frac{a_{n-1}}{a_n}$, $\prod r_i = (-1)^n \frac{a_0}{a_n}$, and everything in between alternates sign.
Where it appears: AIME problems that ask for $\sum r_i^2$, $\sum \frac{1}{r_i}$, or a polynomial with transformed roots — solvable WITHOUT finding the (often irrational) roots. Newton's sums extend Vieta to arbitrary power sums.
Pitfall: watch the leading coefficient — the formulas have $a$ in the denominator, so $r + s = -\frac{b}{a}$, NOT $-b$, unless the polynomial is monic. And mind the alternating signs in higher degree: a dropped sign on $e_2$ or $e_3$ is the classic Vieta error.

Spot the Signal

Look for problems where the key step is connecting polynomial roots to coefficient sums and products to avoid solving explicitly.
You can describe the hard part as connecting polynomial roots to coefficient sums and products to avoid solving explicitly, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for vieta's formulas, then rewrite the givens around it.
Name the relation that makes Vieta's Formulas legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Vieta's Formulas just because the surface looks familiar; verify the required condition first.
Applying Vieta's Formulas because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let vieta's formulas reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is connecting polynomial roots to coefficient sums and products to avoid solving explicitly.