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Module 2 · L6 of 15 ~35 min ⚡ +95 XP available

Sums and Products of Roots — Quadratics

A cryptographer designs quadratic equations whose roots must satisfy hidden constraints. A physicist needs the sum and product of two frequencies without computing them individually. Vieta's formulas — named after the 16th-century mathematician François Viète — are the hidden bridge between a polynomial's coefficients and its roots. They let you answer questions about roots without ever solving the equation. In this lesson you'll derive and master them for quadratics.

Today's hook — For $x^2 - 5x + 6 = 0$, the roots are 2 and 3. Their sum is 5 and their product is 6. Notice: $5 = -(-5)/1$ (the $-b/a$ coefficient) and $6 = 6/1$ (the $c/a$ coefficient). Is this a coincidence? By the end of this lesson you'll know it's always true — and you'll be able to find sums and products of roots for any quadratic in seconds, without solving it.

0/5QUESTS

Recall — your gut answer first

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For $x^2 - 5x + 6 = 0$, the roots are 2 and 3. What is their sum? What is their product? Now look at the coefficients $-5$ and $6$. How do the sum and product of the roots relate to these coefficients? Write your observation before reading on.

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The two moves

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Every Vieta's formulas question in this module uses exactly two ideas. Lock read off $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$ directly from coefficients and use symmetric identities to find related expressions into memory and you'll handle any question fast.

Every Vieta's question lives on one of two roads: read sum and product from coefficients immediately (no solving required), or use identities like $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ to find symmetric expressions from the sum and product.

$\alpha + \beta = -\dfrac{b}{a}, \quad \alpha\beta = \dfrac{c}{a}$

The negative sign matters

$\alpha + \beta = -b/a$, NOT $b/a$. The most common error is forgetting the negative. Write it as "negative $b$ over $a$" every time.

Symmetric identities

$\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ and $\tfrac{1}{\alpha}+\tfrac{1}{\beta} = \tfrac{\alpha+\beta}{\alpha\beta}$. These express derived quantities in terms of sum $S$ and product $P$ only.

Forming new equations

To form a monic quadratic with roots $\alpha$ and $\beta$: $x^2 - (\alpha+\beta)x + \alpha\beta = 0$. Sum becomes coefficient of $x$ (with a minus), product becomes constant.

What you'll master

Know

Key facts

Vieta's formulas for quadratics: $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$
The derivation from expanding $a(x-\alpha)(x-\beta) = ax^2 + bx + c$
Symmetric identities: $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ and $\tfrac{1}{\alpha}+\tfrac{1}{\beta} = \tfrac{\alpha+\beta}{\alpha\beta}$

Understand

Concepts

Why the sum has a negative sign: it comes from expanding $(x-\alpha)(x-\beta)$
Why these formulas work even when the roots are irrational or complex
How to express any symmetric polynomial in terms of $\alpha+\beta$ and $\alpha\beta$

Can do

Skills

Find $\alpha+\beta$ and $\alpha\beta$ directly from coefficients without solving
Evaluate expressions like $\alpha^2+\beta^2$, $\tfrac{1}{\alpha}+\tfrac{1}{\beta}$, $(\alpha-\beta)^2$ using identities
Form a new monic quadratic given the sum and product of its roots

Key terms

Vieta's formulasRelations between the coefficients of a polynomial and the sums and products of its roots, named after François Viète (1540–1603).

Sum of roots $\alpha + \beta$For $ax^2 + bx + c = 0$: $\alpha + \beta = -\dfrac{b}{a}$. Note the negative sign — it comes from the expansion of $a(x-\alpha)(x-\beta)$.

Product of roots $\alpha\beta$For $ax^2 + bx + c = 0$: $\alpha\beta = \dfrac{c}{a}$. The product has no sign change.

Monic quadraticA quadratic with leading coefficient 1, i.e., $x^2 + bx + c$. For monic quadratics, $\alpha + \beta = -b$ and $\alpha\beta = c$.

Symmetric polynomialAn expression in $\alpha$ and $\beta$ that is unchanged when $\alpha$ and $\beta$ are swapped. Examples: $\alpha+\beta$, $\alpha\beta$, $\alpha^2+\beta^2$. All can be expressed using Vieta's formulas.

Forming an equationGiven roots $\alpha$ and $\beta$, the monic quadratic is $x^2 - (\alpha+\beta)x + \alpha\beta = 0$. Multiply through by $a$ if a non-monic form is required.

Vieta's formulas for quadratics — the derivation

core concept

Suppose $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$ (with $a \ne 0$). Then:

$$ax^2 + bx + c = a(x - \alpha)(x - \beta)$$

Expanding the right side:

$$a(x-\alpha)(x-\beta) = a\bigl[x^2 - (\alpha+\beta)x + \alpha\beta\bigr] = ax^2 - a(\alpha+\beta)x + a\alpha\beta$$

Comparing coefficients of $x$ and the constant term with the original $ax^2 + bx + c$:

Coefficient of $x$: $-a(\alpha + \beta) = b$, so $\alpha + \beta = -\dfrac{b}{a}$
Constant term: $a\alpha\beta = c$, so $\alpha\beta = \dfrac{c}{a}$

$$\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}$$

Key insight: These formulas hold for all quadratics, regardless of whether the roots are rational, irrational, or complex. The derivation only requires that $\alpha$ and $\beta$ are the two roots — it never requires you to actually find them.

Why this is powerful. The quadratic formula gives roots of $x^2 - 4x + 2 = 0$ as $x = 2 \pm \sqrt{2}$ — messy numbers. But Vieta's formulas give $\alpha + \beta = 4$ and $\alpha\beta = 2$ instantly from the coefficients, without any computation. Many exam questions ask only for these symmetric quantities — so you never need to find the ugly individual roots.

For ax^2 + bx + c = 0 with roots , : \; + = -b/a and = c/a; Derivation: expand a(x-)(x-) and compare coefficients — the key is the negative in front of (+)

Pause — copy Vieta's formulas for quadratics into your book: for $ax^2 + bx + c = 0$, $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$; derivation — expand $a(x-\alpha)(x-\beta)$ and compare coefficients.

Quick check: For $3x^2 - 7x + 2 = 0$ with roots $\alpha$ and $\beta$, which is correct?

Using symmetric identities

core concept

We just saw that for $ax^2 + bx + c = 0$, the roots satisfy $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$. That raises a question: exam questions often ask for $\alpha^2 + \beta^2$ or $\frac{1}{\alpha} + \frac{1}{\beta}$, not just the sum and product — how do you evaluate these without finding $\alpha$ and $\beta$ individually? This card answers it → rewrite using symmetric identities: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = S^2 - 2P$.

Vieta's formulas give us $\alpha + \beta$ and $\alpha\beta$. Many exam questions ask for other symmetric expressions. The key is to never find $\alpha$ and $\beta$ individually — instead, rewrite the expression using identities:

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$$

$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$$

Strategy: Call the sum $S = \alpha + \beta$ and the product $P = \alpha\beta$. Every symmetric expression can be written in terms of $S$ and $P$ only — no individual roots needed. Then substitute $S = -b/a$ and $P = c/a$.

Forming a new equation: If you know the sum and product of the desired roots, the monic quadratic is immediately:

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0 \qquad \text{i.e.} \quad x^2 - Sx + P = 0$$

^2 + ^2 = (+)^2 - 2 = S^2 - 2P; 1{} + 1{} = +{} = S{P} (only valid when P 0)

Pause — copy the symmetric identity pair into your book: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = S^2 - 2P$; $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{S}{P}$ (only valid when $P \ne 0$).

Did you get this? True or false: for $x^2 - 4x + 2 = 0$ with roots $\alpha$ and $\beta$, $\alpha^2 + \beta^2 = 12$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · READING VIETA'S DIRECTLY

For $2x^2 - 6x + 3 = 0$ with roots $\alpha$ and $\beta$, find $\alpha + \beta$ and $\alpha\beta$.

Identify: $a = 2$, $b = -6$, $c = 3$

Write down the coefficients clearly before applying the formulas. Be careful to include the signs.

PROBLEM 2 · SYMMETRIC IDENTITY

If $\alpha$ and $\beta$ are roots of $x^2 - 4x + 2 = 0$, find $\alpha^2 + \beta^2$.

$\alpha + \beta = 4$, $\alpha\beta = 2$ (monic, so $S = -b = 4$ and $P = c = 2$)

Read off Vieta's formulas directly. For a monic quadratic $x^2 + bx + c$, the sum is $-b$ and the product is $c$.

PROBLEM 3 · FORMING A NEW EQUATION

Form the monic quadratic equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$, given that $\alpha$ and $\beta$ are roots of $x^2 - 3x + 1 = 0$.

From $x^2 - 3x + 1 = 0$: $\alpha + \beta = 3$ and $\alpha\beta = 1$

Read Vieta's formulas for the original equation. We need sum and product of the new roots $1/\alpha$ and $1/\beta$.

Fill the gap: For $x^2 - 5x + 6 = 0$, the sum of roots is and the product of roots is .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the negative sign in the sum

Writing $\alpha + \beta = b/a$ instead of $-b/a$ is the most common Vieta's error. For $x^2 - 5x + 6 = 0$, the sum is $-(-5)/1 = 5$, NOT $-5$. Always write out: "sum $= -b/a$", including the negative, before substituting. The negative comes from the expansion of $(x-\alpha)(x-\beta) = x^2 - (\alpha+\beta)x + \alpha\beta$.

Trap 02

Trying to find the roots individually

Many students use the quadratic formula to find $\alpha$ and $\beta$, then compute the required expression. This is valid but slow — and with irrational roots like $2 \pm \sqrt{2}$, it introduces errors. Vieta's formulas and symmetric identities are always faster and exact. If you find yourself solving the quadratic, ask: "Can I do this using sum and product only?"

Trap 03

Wrong sign when forming the new equation

The monic quadratic is $x^2 - (\alpha+\beta)x + \alpha\beta = 0$. Students often write $x^2 + (\alpha+\beta)x + \alpha\beta = 0$ (positive coefficient of $x$). Remember: the sum of roots becomes the negative coefficient of $x$. If sum $= 3$ and product $= 5$, the equation is $x^2 - 3x + 5 = 0$, not $x^2 + 3x + 5 = 0$.

Did you get this? True or false: the monic quadratic with roots 4 and $-3$ is $x^2 - x - 12 = 0$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

For $3x^2 + 2x - 5 = 0$, find the sum and product of the roots without solving the equation.

If $\alpha$ and $\beta$ are roots of $x^2 - 3x + 1 = 0$, find $\alpha^2 + \beta^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$.

Form the monic quadratic equation with roots 4 and $-3$. Show how you find sum and product first.

If $\alpha$ and $\beta$ are roots of $x^2 - 5x + 3 = 0$, find $(\alpha - \beta)^2$ using the identity $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$.

Explain in your own words why Vieta's formulas are useful. When would you choose them over the quadratic formula?

Revisit your thinking

Earlier you observed that for $x^2 - 5x + 6 = 0$ with roots 2 and 3: sum $= 5 = -(-5)/1$ and product $= 6 = 6/1$. You've now proved this is always true — it follows directly from expanding $a(x-\alpha)(x-\beta)$ and comparing coefficients. The pattern isn't a coincidence; it's a theorem. And it works even when the roots are irrational, such as $\alpha = 2+\sqrt{2}$ and $\beta = 2-\sqrt{2}$: sum $= 4$, product $= 2$ — you can read both from $x^2 - 4x + 2 = 0$ immediately.

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Odd one out: For a quadratic $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, which of the following expressions CANNOT be evaluated using only Vieta's formulas (without finding the individual roots)?

Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. For $3x^2 + 2x - 5 = 0$, find the sum and product of the roots. (2 marks)

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ApplyBand 4–53 marks

Q2. If $\alpha$ and $\beta$ are roots of $x^2 - 3x + 1 = 0$, find $\alpha^2 + \beta^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$. (3 marks)

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ApplyBand 42 marks

Q3. Form the monic quadratic equation with roots 4 and $-3$. (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $a=3, b=2, c=-5$. Sum $= -2/3$, product $= -5/3$. 2. $S=3, P=1$. $\alpha^2+\beta^2 = 9-2 = 7$. $1/\alpha+1/\beta = 3/1 = 3$. 3. Sum $= 1$, product $= -12$. Equation: $x^2 - x - 12 = 0$. 4. $S=5, P=3$. $(\alpha-\beta)^2 = 25-12 = 13$. 5. Vieta's formulas are useful when you need the sum/product/symmetric expressions of roots without actually solving — especially when roots are irrational. Use instead of the quadratic formula whenever the question asks for $\alpha+\beta$, $\alpha\beta$, $\alpha^2+\beta^2$, etc.

Q1 (2 marks): $a=3, b=2, c=-5$ [0.5]. Sum $= -2/3$ [0.75]. Product $= -5/3$ [0.75].

Q2 (3 marks): $S = 3$, $P = 1$ from Vieta's [0.5]. $\alpha^2+\beta^2 = 3^2 - 2(1) = 7$ [1]. $\tfrac{1}{\alpha}+\tfrac{1}{\beta} = \tfrac{S}{P} = 3$ [1.5].

Q3 (2 marks): Sum $= 4 + (-3) = 1$ [0.5]. Product $= 4 \times (-3) = -12$ [0.5]. Equation: $x^2 - x - 12 = 0$ [1].

Boss battle · The Coefficient Detective

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering Vieta's formulas questions. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 5 · Finding Zeros of Polynomials Lesson 7 · Sums and Products of Roots — Cubics →

Module overview · Extension 1 · Checkpoint 2 · Module quiz