Ang Teorema ni Vieta ay nag-uugnay sa mga ugat ng isang quadratic equation sa mga coefficient nito: para sa isang reduced equation na \(x^2 + px + q = 0\), ang suma ng mga ugat ay \(-p\), at ang produkto ay \(q\). Ito ay nagbibigay-daan sa iyo na mahanap ang mga ugat nang pasalita - sa pamamagitan ng pagpili, nang walang discriminant. Sa pahinang ito - ang mga eksaktong pormula para sa reduced equation at para sa pangkalahatang anyo, ang kabaligtaran na teorema ni Vieta, ang algorithm para sa pagpili ng mga ugat, mga halimbawang nasuri, at isang interactive na trainer.
Ano ang Teorema ni Vieta
Ang Teorema ni Vieta ay isang tuntunin na nag-uugnay sa mga ugat ng isang quadratic equation sa mga coefficient nito. Ito ay nakasaad tulad ng sumusunod: kung ang \(x_1\) at \(x_2\) ay ang mga ugat ng reduced quadratic equation na \(x^2 + px + q = 0\), kung gayon
Sa salita: ang suma ng mga ugat ay katumbas ng pangalawang coefficient na may kabaligtaran na tanda, at ang produkto ng mga ugat ay katumbas ng constant term. Ang minus sign sa suma ay ang pinakamahalagang bahagi ng buong paksa: ito ay \(-p\), hindi \(p\).
Para sa pangkalahatang anyo ng equation na \(ax^2 + bx + c = 0\) (kung saan \(a \neq 0\)), ang mga pormula ay nakukuha sa pamamagitan ng paghahati sa \(a\):
Ang teorema ay gumagana lamang kapag mayroong mga tunay na ugat, ibig sabihin, kapag \(D \ge 0\). Kung ang discriminant ay negatibo, walang mapipili: walang mga tunay na numero na may ganitong suma at ganitong produkto.
Reduced Equation and General Form: Where to Divide by a
A quadratic equation is called reduced if the coefficient of \(x^2\) is one: \(x^2 + px + q = 0\). It is for this equation that Vieta's theorem is stated most concisely, which is why the selection of roots always begins with reduction.
If \(a \neq 1\), the equation is first divided term by term by \(a\). For example, \(3x^2 - 10x + 3 = 0\) after dividing by \(3\) becomes \(x^2 - \frac{10}{3}x + 1 = 0\), i.e., \(p = -\frac{10}{3}\), \(q = 1\). From this, \(x_1 + x_2 = \frac{10}{3}\) and \(x_1 \cdot x_2 = 1\).
The same idea can be written immediately for the general form without reducing the equation: \(x_1 + x_2 = -\frac{b}{a}\), \(x_1 \cdot x_2 = \frac{c}{a}\). The most common mistake is to apply the "short" formulas \(-p\) and \(q\) to an unreduced equation and get a sum of \(10\) instead of \(\frac{10}{3}\).
Converse of Vieta's Theorem
The Vieta's theorem has a converse, and it is this that allows equations to be solved by selection. It states that 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 already known, we get the relationships between them. The converse goes from coefficients to roots: we guess a pair of numbers with the required sum and product - and we have every right to call them roots, nothing more needs to be proven.
From this also follows 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, \(x^2 - 5x + 6 = (x-2)(x-3)\).
Algorithm for Selecting Roots Using Vieta's Theorem
Selecting roots is the main practical application of the theorem. It works faster than the discriminant when the roots are integers, and takes five steps.
Signs of Roots Based on Coefficients p and q
Even before selecting, you can determine the signs of the roots by looking at the signs of \(p\) and \(q\). This significantly reduces the search: you can immediately see whether to look for a pair of two positive numbers, two negative numbers, or numbers with different signs.
| p at q | Mga Ugat | Halimbawa |
|---|---|---|
| q > 0, p < 0 | parehong > 0 | x2 − 5x + 6 = 0 → 2; 3 |
| q > 0, p > 0 | parehong < 0 | x2 + 5x + 6 = 0 → −2; −3 |
| q < 0 | magkaibang tanda | x2 + 2x − 15 = 0 → 3; −5 |
| q = 0 | 0 at −p | x2 − 5x = 0 → 0; 5 |
Vieta's Theorem or Discriminant: Which to Choose
The formula for roots using the discriminant \(x = \dfrac{-b \pm \sqrt{D}}{2a}\) always works but requires calculations. Vieta's theorem works instantly, but only if the roots are guessable - usually small integers.
The practical rule is as follows:
- the equation is reduced, and \(q\) can be factored into small factors → try Vieta's selection, the answer is obtained mentally;
- the coefficients are large or fractional, the roots are not guessable within 10-15 seconds → calculate the discriminant;
- only the number of roots or the signs of the roots are needed, and the roots themselves are not → \(D\) and the sign table above are sufficient.
It is also useful to combine: calculate the discriminant and find the roots → check the answer using Vieta's theorem. The sum should give \(-p\), the product \(q\). Such a check takes a couple of seconds and catches almost any arithmetic error.
Examples of Solving Using Vieta's Theorem
Example 1. Selecting Roots: Both Roots Are Positive
Solve \(x^2 - 9x + 20 = 0\).
Step 1. The equation is reduced: \(p = -9\), \(q = 20\).
Step 2. We check the discriminant: \(D = (-9)^2 - 4\cdot 20 = 81 - 80 = 1 > 0\) - there are two roots.
Step 3. We write down the relations: \(x_1 + x_2 = -p = 9\), \(x_1 \cdot x_2 = q = 20\).
Step 4. We go through the factorizations 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. Check 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 reduced: \(p = 3\), \(q = -28\).
Step 2. \(q < 0\), which means 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. Relations: \(x_1 + x_2 = -3\), \(x_1 \cdot x_2 = -28\).
Step 4. Factorizations 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, we take the root with the larger absolute value with a minus sign: \(4\) and \(-7\).
Step 5. Check. \(4^2 + 3\cdot 4 - 28 = 16 + 12 - 28 = 0\); \((-7)^2 + 3\cdot(-7) - 28 = 49 - 21 - 28 = 0\).
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 applied. We 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 - this means the roots are reciprocals: if one 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. Check 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 reduced quadratic equation whose roots are \(-4\) and \(6\).
Step 1. We 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. We write the equation: \(x^2 - 2x - 24 = 0\).
Step 4. Check 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 by factoring \((x + 4)(x - 6) = 0\).
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: a sign was missed.
Step 3. We look for 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. Check 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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Losing the minus sign in the sum: writing $x_1 + x_2 = p$ instead of $x_1 + x_2 = -p$.
The minus sign is only for the sum, the product is taken with the same sign: \(x_1 + x_2 = -p\), \(x_1 x_2 = q\). Check with a 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 an unreduced equation: for $2x^2 - 9x + 4 = 0$, saying "sum is 9, product is 4".
As long as \(a \neq 1\), the general form formulas apply: \(x_1 + x_2 = -\dfrac{b}{a} = \dfrac{9}{2}\), \(x_1 x_2 = \dfrac{c}{a} = 2\). The roots here are 4 and \(\tfrac{1}{2}\), not 4 and 1. Or, first divide the entire equation by \(a\).
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Selecting roots without checking the discriminant: for $x^2 + x + 1 = 0$, trying to find a pair of numbers with a sum of $-1$ and a product of $1$.
First, make sure that roots exist: \(D = 1 - 4 = -3 < 0\), there are no real roots, so selection is impossible. Vieta's theorem works only when \(D \ge 0\).
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Selecting a pair based only on the product, without checking the sum: for $x^2 - 8x + 12 = 0$, taking 3 and 4 because their product is 12.
Both relations 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 given only 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, taking $p$ 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 \(x^2 - 7x + 10 = 0\). 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 relations \(x_1 + x_2 = -p\) and \(x_1 x_2 = q\) hold for any real roots - integer, fractional, and irrational. Simple mental selection 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 reduced quadratic equation \(x^2 + px + q = 0\) has roots, then their sum is equal to the second coefficient with the opposite sign, and their product is equal to 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 the 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 reduced 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 get the sum \(-p\) and the product \(q\). The converse goes the other way: if a pair of numbers gives the required sum and product, then these numbers are the roots of the equation. It is the converse theorem that allows equations to be solved by selection.
How to select roots using Vieta's theorem?
Reduce the equation to the form \(x^2 + px + q = 0\), make sure that \(D \ge 0\), write down \(x_1 + x_2 = -p\) and \(x_1 x_2 = q\), go through the factorizations of \(q\) into pairs of factors and choose the pair whose sum is equal to \(-p\). 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 valid only for a reduced equation, so when \(a \neq 1\), the equation is first divided by \(a\).
What to do if the roots cannot be selected using Vieta's theorem?
This means the roots are not integers or do not exist at all. Calculate the discriminant: if \(D < 0\), there are no real roots, and if \(D \ge 0\), find the roots using the formula \(x = \dfrac{-b \pm \sqrt{D}}{2a}\) and then check the answer using Vieta's theorem.
How to 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 \(q < 0\) - roots have different signs; if \(q = 0\) - one of the roots is zero.
In what 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 a parameter, and in OGE and EGE tasks.
