Vieta's theorem relates the roots of a quadratic equation to its coefficients: for a monic equation \(x^2 + px + q = 0\), the sum of the roots is \(-p\), and the product is \(q\). This allows for mental calculation of roots, without using the discriminant. This page covers the exact formulas for monic and general forms, the converse of Vieta's theorem, an algorithm for finding roots by inspection, solved examples, and an interactive trainer.
What is Vieta's Theorem
Vieta's theorem is a rule that relates the roots of a quadratic equation to its coefficients. It is stated as follows: if \(x_1\) and \(x_2\) are the roots of the monic quadratic equation \(x^2 + px + q = 0\), then
In words: the sum of the roots equals the second coefficient with the opposite sign, and the product of the roots equals the constant term. The minus sign in the sum is the most crucial part of the entire topic: it must be \(-p\), not \(p\).
For a general form equation \(ax^2 + bx + c = 0\) (where \(a \neq 0\)), the formulas are obtained by dividing by \(a\):
The theorem works only when real roots exist, i.e., when \(D \ge 0\). If the discriminant is negative, there's nothing to find: no real numbers exist with such a sum and product.
Monic and General Form: When to Divide by a
A quadratic equation is called monic if the coefficient of \(x^2\) is one: \(x^2 + px + q = 0\). Vieta's theorem is stated most concisely for this form, which is why finding roots by inspection always starts with converting to the monic form.
If \(a \neq 1\), the equation is first divided term by term by \(a\). For example, \(3x^2 - 10x + 3 = 0\) becomes \(x^2 - \frac{10}{3}x + 1 = 0\) after dividing by \(3\), meaning \(p = -\frac{10}{3}\) and \(q = 1\). Thus, \(x_1 + x_2 = \frac{10}{3}\) and \(x_1 \cdot x_2 = 1\).
The same idea can be written directly for the general form without converting to monic: \(x_1 + x_2 = -\frac{b}{a}\), \(x_1 \cdot x_2 = \frac{c}{a}\). The most common mistake is applying the 'short' formulas \(-p\) and \(q\) to a non-monic equation, resulting in a sum of \(10\) instead of \(\frac{10}{3}\).
Converse of Vieta's Theorem
Vieta's theorem has a converse, which is what allows us to solve equations by inspection. It states: if numbers \(m\) and \(n\) are such that
then \(m\) and \(n\) are the roots of the equation \(x^2 + px + q = 0\).
The difference is fundamental. The direct theorem goes from roots to coefficients: the roots are known, and we find the relationships between them. The converse goes from coefficients to roots: if we guess a pair of numbers with the required sum and product, we are fully entitled to call them roots; no further proof is needed.
This also leads to a useful rule: if \(x_1\) and \(x_2\) are the roots, then the polynomial can be factored as \(x^2 + px + q = (x - x_1)(x - x_2)\). For example, \(x^2 - 5x + 6 = (x-2)(x-3)\).
Finding Roots by Inspection Using Vieta's Theorem: Algorithm
Finding roots by inspection is the main practical application of the theorem. It's faster than using the discriminant when the roots are integers and involves five steps.
Signs of Roots Based on Coefficients p and q
Even before attempting to find the roots, you can determine their signs by looking at the signs of \(p\) and \(q\). This significantly reduces the number of possibilities: you'll immediately know whether to look for a pair of positive numbers, a pair of negative numbers, or numbers with different signs.
| p og q | Røtter | Eksempel |
|---|---|---|
| q > 0, p < 0 | begge > 0 | x2 − 5x + 6 = 0 → 2; 3 |
| q > 0, p > 0 | begge < 0 | x2 + 5x + 6 = 0 → −2; −3 |
| q < 0 | ulike fortegn | x2 + 2x − 15 = 0 → 3; −5 |
| q = 0 | 0 og −p | x2 − 5x = 0 → 0; 5 |
Vieta's Theorem vs. Discriminant: Which to Choose
The quadratic formula \(x = \dfrac{-b \pm \sqrt{D}}{2a}\) always works but requires calculations. Vieta's theorem works instantly but only if the roots can be guessed, which is usually the case for small integers.
Here's a practical rule:
- The equation is monic, and \(q\) can be factored into small integers → try finding roots by inspection using Vieta's theorem; the answer is obtained mentally.
- The coefficients are large or fractional, and the roots are not obvious within 10–15 seconds → calculate the discriminant.
- You only need the number of roots or their signs, not the roots themselves → the discriminant and the sign table above are sufficient.
It's also useful to combine methods: after calculating the discriminant and finding the roots, check your answer using Vieta's theorem. The sum should be \(-p\), and the product should be \(q\). This check takes a couple of seconds and catches almost any arithmetic error.
Examples of Solving Using Vieta's Theorem
Example 1. Finding Roots by Inspection: Both Roots Are Positive
Solve \(x^2 - 9x + 20 = 0\).
Step 1. The equation is monic: \(p = -9\), \(q = 20\).
Step 2. Check the discriminant: \(D = (-9)^2 - 4\cdot 20 = 81 - 80 = 1 > 0\) — there are two roots.
Step 3. Write down the relationships: \(x_1 + x_2 = -p = 9\), \(x_1 \cdot x_2 = q = 20\).
Step 4. Consider the factor pairs of 20: \(1\cdot 20\) (sum 21), \(2\cdot 10\) (sum 12), \(4\cdot 5\) (sum 9) — the last pair fits. Both factors are positive because \(q > 0\) and \(p < 0\).
Step 5. Verify by substitution. \(4^2 - 9\cdot 4 + 20 = 16 - 36 + 20 = 0\); \(5^2 - 9\cdot 5 + 20 = 25 - 45 + 20 = 0\).
Answer: \(x_1 = 4\), \(x_2 = 5\).
Example 2. Roots of Different Signs
Solve \(x^2 + 3x - 28 = 0\).
Step 1. The equation is monic: \(p = 3\), \(q = -28\).
Step 2. Since \(q < 0\), the roots have different signs, and the discriminant is guaranteed to be positive: \(D = 3^2 - 4\cdot(-28) = 9 + 112 = 121 > 0\).
Step 3. Relationships: \(x_1 + x_2 = -3\), \(x_1 \cdot x_2 = -28\).
Step 4. Factor pairs of 28: \(1\cdot 28\), \(2\cdot 14\), \(4\cdot 7\). A difference of 3 is needed — the pair 4 and 7 fits. Since the sum is negative, the root with the larger absolute value is negative: \(4\) and \(-7\).
Step 5. Verification. \(4^2 + 3\cdot 4 - 28 = 16 + 12 - 28 = 0\); \((-7)^2 + 3\cdot(-7) - 28 = 49 - 21 - 28 = 0\).
Answer: \(x_1 = 4\), \(x_2 = -7\).
Example 3. General Form Equation (a ≠ 1)
Solve \(3x^2 - 10x + 3 = 0\).
Step 1. The coefficient \(a = 3\), so the 'short' formulas \(-p\) and \(q\) cannot be directly applied. Use the general form: \(x_1 + x_2 = -\dfrac{b}{a} = \dfrac{10}{3}\), \(x_1 \cdot x_2 = \dfrac{c}{a} = \dfrac{3}{3} = 1\).
Step 2. The product is 1, meaning the roots are reciprocals: if one root is \(t\), the other is \(\tfrac{1}{t}\). The sum \(t + \tfrac{1}{t} = \tfrac{10}{3}\) suggests \(t = 3\), since \(3 + \tfrac{1}{3} = \tfrac{10}{3}\).
Step 3. Verify by substitution. \(3\cdot 3^2 - 10\cdot 3 + 3 = 27 - 30 + 3 = 0\); \(3\cdot\left(\tfrac{1}{3}\right)^2 - 10\cdot\tfrac{1}{3} + 3 = \tfrac{1}{3} - \tfrac{10}{3} + 3 = 0\).
Answer: \(x_1 = 3\), \(x_2 = \tfrac{1}{3}\).
Example 4. Constructing an Equation from its Roots
Task. Construct a monic quadratic equation whose roots are \(-4\) and \(6\).
Step 1. Calculate the sum and product: \(x_1 + x_2 = -4 + 6 = 2\), \(x_1 \cdot x_2 = -4 \cdot 6 = -24\).
Step 2. By the converse of Vieta's theorem, the coefficients are \(p = -(x_1 + x_2) = -2\) and \(q = x_1 x_2 = -24\). Here lies the main trap: \(p\) is the sum with the opposite sign.
Step 3. Write the equation: \(x^2 - 2x - 24 = 0\).
Step 4. Verify by substitution. \((-4)^2 - 2\cdot(-4) - 24 = 16 + 8 - 24 = 0\); \(6^2 - 2\cdot 6 - 24 = 36 - 12 - 24 = 0\).
Answer: \(x^2 - 2x - 24 = 0\). The same result is obtained from the factorization \((x + 4)(x - 6) = 0\).
Example 5. Checking Someone Else's Answer Using Vieta's Theorem
Task. A student solved the equation \(x^2 + 5x + 6 = 0\) and wrote the answer: \(x_1 = 2\), \(x_2 = 3\). Check if they made a mistake.
Step 1. According to Vieta's theorem, we should have \(x_1 + x_2 = -p = -5\) and \(x_1 \cdot x_2 = q = 6\).
Step 2. The proposed answer's product matches (\(2\cdot 3 = 6\)), but the sum does not: \(2 + 3 = 5 \neq -5\). Therefore, the answer is incorrect: the sign was missed.
Step 3. Find the correct pair: we need numbers with a product of \(6\) and a sum of \(-5\) — these are \(-2\) and \(-3\) (both negative because \(q > 0\) and \(p > 0\)).
Step 4. Verify by substitution. \((-2)^2 + 5\cdot(-2) + 6 = 4 - 10 + 6 = 0\); \((-3)^2 + 5\cdot(-3) + 6 = 9 - 15 + 6 = 0\).
Answer: The correct roots are \(x_1 = -2\), \(x_2 = -3\).
Common Mistakes When Working with Vieta's Theorem
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Forgetting the minus sign in the sum: writing $x_1 + x_2 = p$ instead of $x_1 + x_2 = -p$.
The minus sign only applies to the sum; the product keeps its original sign: \(x_1 + x_2 = -p\), \(x_1 x_2 = q\). Check with the simple equation \(x^2 - 5x + 6 = 0\): here \(p = -5\), the sum of the roots is \(5\), and the roots are indeed 2 and 3.
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Applying the formulas $-p$ and $q$ to a non-monic equation: for $2x^2 - 9x + 4 = 0$, stating 'sum is 9, product is 4'.
As long as \(a \neq 1\), use the general form formulas: \(x_1 + x_2 = -\dfrac{b}{a} = \dfrac{9}{2}\), \(x_1 x_2 = \dfrac{c}{a} = 2\). The roots here are 4 and \(\tfrac{1}{2}\), not 4 and 1. Alternatively, divide the entire equation by \(a\) first.
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Trying to find roots by inspection without checking the discriminant: for $x^2 + x + 1 = 0$, attempting to find a pair of numbers with a sum of $-1$ and a product of $1$.
First, ensure that roots exist: \(D = 1 - 4 = -3 < 0\), there are no real roots, so inspection is impossible. Vieta's theorem only works when \(D \ge 0\).
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Finding a factor pair for the product only, without checking the sum: for $x^2 - 8x + 12 = 0$, choosing 3 and 4 because their product is 12.
Both relationships must be checked simultaneously. For \(x^2 - 8x + 12 = 0\), the product 12 is given by pairs (1; 12), (2; 6), (3; 4), but the sum 8 is only given by the pair 2 and 6. Check: \(2^2 - 8\cdot 2 + 12 = 0\) and \(6^2 - 8\cdot 6 + 12 = 0\).
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When constructing an equation from roots, using $p$ equal to the sum: for roots 2 and 5, writing $x^2 + 7x + 10 = 0$.
The coefficient \(p\) is equal to the sum with the opposite sign: \(p = -(2 + 5) = -7\), \(q = 2\cdot 5 = 10\), so the equation is \(x^2 - 7x + 10 = 0\). Substitution confirms: \(2^2 - 7\cdot 2 + 10 = 0\), \(5^2 - 7\cdot 5 + 10 = 0\).
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Assuming Vieta's theorem is only valid for integer roots and not using it to check answers.
The relationships \(x_1 + x_2 = -p\) and \(x_1 x_2 = q\) hold for any real roots — integer, fractional, or irrational. Mental inspection is convenient for integers, but for others, the theorem remains a quick way to check an answer found using the discriminant.
Questions and Answers
How is Vieta's theorem stated in simple terms?
If a monic quadratic equation \(x^2 + px + q = 0\) has roots, then their sum equals the second coefficient with the opposite sign, and their product equals the constant term: \(x_1 + x_2 = -p\) and \(x_1 x_2 = q\).
What are the sum and product of the roots of the equation $ax^2 + bx + c = 0$?
For a general form equation, the sum of the roots is \(-\dfrac{b}{a}\), and the product is \(\dfrac{c}{a}\). These formulas are derived from the monic case by dividing the entire equation by \(a\).
How does the converse of Vieta's theorem differ from the direct theorem?
The direct theorem goes from roots to coefficients: knowing the roots, we find the sum \(-p\) and product \(q\). The converse goes the other way: if a pair of numbers yields the required sum and product, then those numbers are the roots of the equation. It is the converse theorem that allows solving equations by inspection.
How do you find roots by inspection using Vieta's theorem?
Convert the equation to the form \(x^2 + px + q = 0\), ensure that \(D \ge 0\), write down \(x_1 + x_2 = -p\) and \(x_1 x_2 = q\), consider factor pairs of \(q\), and choose the pair whose sum equals \(-p\). For example, for \(x^2 + 6x + 8 = 0\), we need numbers with a product of 8 and a sum of \(-6\) — these are \(-2\) and \(-4\).
Does Vieta's theorem work if the coefficient of $x^2$ is not one?
Yes, but in the general form: \(x_1 + x_2 = -\dfrac{b}{a}\), \(x_1 x_2 = \dfrac{c}{a}\). The short formulas \(-p\) and \(q\) are only valid for monic equations, so if \(a \neq 1\), the equation must first be divided by \(a\).
What if the roots cannot be found by inspection using Vieta's theorem?
This means the roots are either not integers or do not exist. Calculate the discriminant: if \(D < 0\), there are no real roots; if \(D \ge 0\), find the roots using the formula \(x = \dfrac{-b \pm \sqrt{D}}{2a}\) and then check your answer using Vieta's theorem.
How can you determine the signs of the roots without solving the equation?
By the signs of \(p\) and \(q\) when \(D \ge 0\): if \(q > 0\) and \(p < 0\), both roots are positive; if \(q > 0\) and \(p > 0\), both roots are negative; if \(q < 0\), the roots have different signs; if \(q = 0\), one of the roots is zero.
In which grade is Vieta's theorem taught?
In the 8th grade during algebra lessons, immediately after quadratic equations and the discriminant. It is further used when factoring a quadratic trinomial into factors \(x^2 + px + q = (x - x_1)(x - x_2)\), when analyzing roots with parameters, and in OGE and ЕГЭ tasks.
