韦达定理wéidá dìnglǐ

Core Concept

Vieta's formulas (韦达定理) relate the coefficients of a polynomial to sums and products of its roots.

For the quadratic equation $ax^2 + bx + c = 0$ with roots $x_1$ and $x_2$:

$x_1 + x_2 = -\frac{b}{a}$ (Sum of roots)

$x_1 \cdot x_2 = \frac{c}{a}$ (Product of roots)

Derivation

From the quadratic formula: $x_1 = \frac{-b + \sqrt{\Delta}}{2a}, \quad x_2 = \frac{-b - \sqrt{\Delta}}{2a}$

Sum: $x_1 + x_2 = \frac{-b + \sqrt{\Delta}}{2a} + \frac{-b - \sqrt{\Delta}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}$

Product: $x_1 \cdot x_2 = \frac{(-b + \sqrt{\Delta})(-b - \sqrt{\Delta})}{4a^2} = \frac{b^2 - \Delta}{4a^2} = \frac{b^2 - (b^2 - 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}$

Important Applications

1. Finding Expressions Without Solving

Calculate expressions involving roots without finding the actual roots.

Common formulas: $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2$

$x_1^3 + x_2^3 = (x_1 + x_2)^3 - 3x_1x_2(x_1 + x_2)$

$\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2}$

$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$

2. Constructing Equations from Roots

If $\alpha$ and $\beta$ are roots, then the quadratic equation is: $x^2 - (\alpha + \beta)x + \alpha\beta = 0$

Or equivalently: $(x - \alpha)(x - \beta) = 0$

3. Determining Root Properties

Without solving, determine if roots are:

• Both positive: $x_1 + x_2 > 0$ AND $x_1 \cdot x_2 > 0$
• Both negative: $x_1 + x_2 < 0$ AND $x_1 \cdot x_2 > 0$
• Opposite signs: $x_1 \cdot x_2 < 0$

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus.

Example 1: Basic (Difficulty ★★☆☆☆)

If $x_1$ and $x_2$ are roots of $x^2 - 3x - 4 = 0$, find $x_1 + x_2$ and $x_1 \cdot x_2$.

Solution:

By Vieta's formulas with $a = 1$, $b = -3$, $c = -4$: $x_1 + x_2 = -\frac{-3}{1} = 3$ $x_1 \cdot x_2 = \frac{-4}{1} = -4$

Answer: Sum = 3, Product = -4

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Example 2: Intermediate (Difficulty ★★★☆☆)

If $x_1$ and $x_2$ are roots of $2x^2 + 5x - 3 = 0$, find $x_1^2 + x_2^2$.

Solution:

By Vieta's formulas: $x_1 + x_2 = -\frac{5}{2}, \quad x_1 \cdot x_2 = -\frac{3}{2}$

Using the identity: $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2$ $= \left(-\frac{5}{2}\right)^2 - 2\left(-\frac{3}{2}\right)$ $= \frac{25}{4} + 3 = \frac{25}{4} + \frac{12}{4} = \frac{37}{4}$

Answer: $\dfrac{37}{4}$

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Example 3: Advanced (Difficulty ★★★★☆)

If $x_1$ and $x_2$ are roots of $x^2 - 4x + 1 = 0$, find $\dfrac{x_1}{x_2} + \dfrac{x_2}{x_1}$.

Solution:

By Vieta's formulas: $x_1 + x_2 = 4, \quad x_1 \cdot x_2 = 1$

Simplify the expression: $\frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{x_1^2 + x_2^2}{x_1 x_2}$

First find $x_1^2 + x_2^2$: $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = 16 - 2 = 14$

Therefore: $\frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{14}{1} = 14$

Answer: $14$

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Example 4: Advanced (Difficulty ★★★★☆)

Find a quadratic equation with integer coefficients whose roots are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

Solution:

Sum of roots: $\alpha + \beta = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$

Product of roots: $\alpha \cdot \beta = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$

The equation is: $x^2 - 4x + 1 = 0$

Answer: $x^2 - 4x + 1 = 0$

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Example 5: Advanced (Difficulty ★★★★★)

If one root of $x^2 + px + q = 0$ is double the other, and the sum of roots is 6, find $p$ and $q$.

Solution:

Let roots be $r$ and $2r$.

From Vieta's formulas:

• $r + 2r = 3r = -p$
• $r \cdot 2r = 2r^2 = q$

Given sum = 6: $3r = 6 \Rightarrow r = 2$

Therefore: $p = -3r = -6$ $q = 2r^2 = 2(4) = 8$

Answer: $p = -6$, $q = 8$

Extended Vieta's Formulas

For cubic equation $ax^3 + bx^2 + cx + d = 0$ with roots $x_1, x_2, x_3$:

$x_1 + x_2 + x_3 = -\frac{b}{a}$ $x_1x_2 + x_2x_3 + x_3x_1 = \frac{c}{a}$ $x_1 \cdot x_2 \cdot x_3 = -\frac{d}{a}$

Common Mistakes

❌ Mistake 1: Wrong Sign for Sum

Wrong: $x_1 + x_2 = \frac{b}{a}$ ✗

Correct: $x_1 + x_2 = -\frac{b}{a}$ ✓

❌ Mistake 2: Assuming Roots Are Real

Vieta's formulas work even when $\Delta < 0$ (complex roots), but many applications require real roots. Always check $\Delta \geq 0$ if needed.

❌ Mistake 3: Incomplete Root Conditions

For "both roots positive": need BOTH $x_1 + x_2 > 0$ AND $x_1 \cdot x_2 > 0$ AND $\Delta \geq 0$.

Study Tips

1. ✅ Memorize the sign: Sum has negative, product doesn't
2. ✅ Learn common transformations: $x_1^2 + x_2^2$, $\frac{1}{x_1} + \frac{1}{x_2}$, etc.
3. ✅ Combine with discriminant: For real root conditions
4. ✅ Practice reverse problems: Finding equations from given roots

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💡 Exam Tip: Vieta's formulas are tested frequently in CSCA. Master the transformations for $x_1^2 + x_2^2$ and $\frac{1}{x_1} + \frac{1}{x_2}$ - they appear in most problems!