Worked examples — Vieta's formulas — sum and product of roots

Before anything, let us re-anchor the only two tools we use, in plain words.

Two words we will use repeatedly:

• Leading coefficient $a$ — the number multiplying $x^2$. If it is $1$ the quadratic is called monic.
• Root — a value of $x$ that makes the expression equal zero. A quadratic has exactly two (counting a repeat as two, and allowing imaginary ones).

If you want to know how many real roots exist and whether they repeat, that is a job for the Discriminant; Vieta's formulas work regardless of what the roots turn out to be, which is exactly their charm.

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The scenario matrix

Every problem this topic can throw at you lands in one of these cells. The last column tells you which worked example covers it.

Cell	What makes it tricky	Sign / degenerate feature	Example
A. Clean monic	$a=1$, tidy integers	both roots positive	Ex 1
B. Non-monic	$a\ne 1$, must divide first	mixed signs	Ex 2
C. Negative $b$ and $c$	sign bookkeeping	sum $>0$, product $<0$	Ex 3
D. A root is zero	$c=0$ degenerate	product $=0$	Ex 4
E. Equal roots	discriminant $=0$	$\alpha =\beta$	Ex 5
F. Imaginary roots	no real solutions	sum/product still real	Ex 6
G. Transformed roots	build a new equation	reciprocals / squares	Ex 7
H. Word problem (geometry)	translate then apply	rectangle sides	Ex 8
I. Exam twist	find a hidden coefficient	one root & Vieta given	Ex 9

We will hit all nine cells with nine examples. Read the Forecast line and guess before scrolling.

The figure above is our sign-map: it shows, at a glance, what the signs of the sum and product tell you about the roots — we will lean on it in Examples 3, 4 and 8.

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Cell A — Clean monic quadratic

Forecast: the roots factor as $3$ and $4$ — can you predict the sum and product before computing?

1. Identify $a,b,c$. Here $a=1,b=-7,c=12$. Why this step? Vieta's formulas are stated in terms of these three numbers, so pinning them down (with their signs) is always move one.
2. Sum. $\alpha +\beta =-\frac{b}{a}=-\frac{-7}{1}=7.$ Why this step? Applying the sum rule directly; the double negative flips $-7$ to $+7$.
3. Product. $\alpha \beta =\frac{c}{a}=\frac{12}{1}=12.$ Why this step? The product rule reads the constant term straight off.

Verify: factor $x^2-7x+12=(x-3)(x-4)$, so roots are $3,4$; indeed $3+4=7$ and $3\cdot 4=12$. ✓

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Cell B — Non-monic (divide by $a$ first!)

Forecast: because $c=-5$ is negative, the product will be negative — so the roots have opposite signs. Guess the sum too.

1. Identify. $a=2,b=-3,c=-5$.
2. Sum. $\alpha +\beta =-\frac{b}{a}=-\frac{-3}{2}=\frac{3}{2}.$ Why this step? We must divide by $a=2$; skipping this is the single most common error (see parent Mistake 2).
3. Product. $\alpha \beta =\frac{c}{a}=\frac{-5}{2}=-\frac{5}{2}.$

Verify: actual roots via Quadratic Formula or Factoring Quadratics: $2x^2-3x-5=(2x-5)(x+1)$ gives $x=\frac{5}{2}$ and $x=-1$. Sum $=\frac{5}{2}-1=\frac{3}{2}$ ✓, product $=\frac{5}{2}\cdot (-1)=-\frac{5}{2}$ ✓.

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Cell C — Negative middle, negative constant (sign bookkeeping)

Forecast: the $x^2$ coefficient is negative. Does that change anything? Look at the sign-map figure: product negative $\Rightarrow$ opposite-sign roots.

1. Identify — carefully. $a=-1,b=6,c=16$. Why this step? The leading sign is easy to miss; $a=-1$, not $+1$.
2. Sum. $\alpha +\beta =-\frac{b}{a}=-\frac{6}{-1}=6.$ Why this step? Two minuses again cancel; the answer stays positive.
3. Product. $\alpha \beta =\frac{c}{a}=\frac{16}{-1}=-16.$ Why this step? Dividing a positive by $-1$ flips the sign — product is negative, confirming opposite-sign roots.

Verify: multiply the whole equation by $-1$: $x^2-6x-16=0=(x-8)(x+2)$, roots $8$ and $-2$. Sum $=6$ ✓, product $=-16$ ✓. Multiplying an equation by a constant does not move its roots, so Vieta's answers are identical — a good sanity habit.

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Cell D — A root is zero (degenerate: $c=0$)

Forecast: there is no constant term ($c=0$). What must the product of the roots therefore be?

1. Identify. $a=5,b=-20,c=0$. Why this step? Writing $c=0$ explicitly reminds us the constant term is genuinely there — it's just zero.
2. Product. $\alpha \beta =\frac{c}{a}=\frac{0}{5}=0.$ Why this step? A product being $0$ forces at least one root to be $0$ — the degenerate case the sign-map marks on the boundary line.
3. Sum. $\alpha +\beta =-\frac{b}{a}=-\frac{-20}{5}=4.$ Why this step? Even with a zero root, the sum rule is unchanged; the other root must then be the whole sum.

Verify: $5x^2-20x=5x(x-4)$, roots $0$ and $4$. Sum $=4$ ✓, product $=0$ ✓. The zero root simply "donates" nothing to the sum and kills the product.

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Cell E — Equal roots (discriminant $=0$)

Forecast: this is a perfect square. If both roots are equal, sum and product still exist — what should they be?

1. Identify. $a=1,b=-10,c=25$.
2. Sum. $\alpha +\beta =-\frac{-10}{1}=10.$
3. Product. $\alpha \beta =\frac{25}{1}=25.$
4. Solve for the repeated root. If $\alpha =\beta$, then $2\alpha =10\Rightarrow \alpha =5$, and indeed ${\alpha }^2=25$. Why this step? When roots are equal Vieta's still holds; we exploit $\alpha =\beta$ to split the sum in half and cross-check with the product.

Verify: $x^2-10x+25=(x-5{)}^2$, a double root at $5$. Sum $5+5=10$ ✓, product $5\cdot 5=25$ ✓. (The Discriminant here is $b^2-4ac=100-100=0$, the signature of equal roots.)

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Cell F — Imaginary roots (no real solutions)

Forecast: the discriminant is negative, so the parabola never touches the $x$-axis. Do Vieta's formulas still give real numbers?

1. Identify. $a=1,b=2,c=5$.
2. Sum. $\alpha +\beta =-\frac{2}{1}=-2.$
3. Product. $\alpha \beta =\frac{5}{1}=5.$ Why this step? We applied the rules blind to whether the roots are real — that is the point: Vieta's are algebraic identities, not geometry.

Verify: by Quadratic Formula, $x=\frac{-2\pm \sqrt{4-20}}{2}=-1\pm 2i$. The two roots are complex conjugates. Their sum $(-1+2i)+(-1-2i)=-2$ ✓ (imaginary parts cancel), and product $(-1{)}^2-(2i{)}^2=1+4=5$ ✓. Real coefficients $\Rightarrow$ real sum and product, always.

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Cell G — Transformed roots (build a new equation)

Forecast: the new sum is $\frac{1}{\alpha }+\frac{1}{\beta }$ — can you write that using only $\alpha +\beta$ and $\alpha \beta$? (This is a symmetric expression trick.)

1. Old sum & product. $\alpha +\beta =-\frac{-4}{3}=\frac{4}{3},\alpha \beta =\frac{1}{3}.$
2. New sum. $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{4/3}{1/3}=4.$ Why this step? Combining the reciprocals over a common denominator turns them into the ratio of the two quantities we already know — no need to find $\alpha ,\beta$ individually.
3. New product. $\frac{1}{\alpha }\cdot \frac{1}{\beta }=\frac{1}{\alpha \beta }=\frac{1}{1/3}=3.$
4. Assemble using $x^2-(\text{sum})x+(\text{product})=0$: $x^2-4x+3=0.$

Verify: old roots of $3x^2-4x+1=0=(3x-1)(x-1)$ are $\frac{1}{3}$ and $1$; their reciprocals are $3$ and $1$. Check $x^2-4x+3=(x-3)(x-1)$ — roots $3,1$ ✓.

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Cell H — Word problem (geometry)

Forecast: the two sides are the two roots of one quadratic. What are their sum and product? Glance at the sign-map: both quantities positive $\Rightarrow$ both roots positive (lengths must be!).

1. Translate the words. Perimeter $26\Rightarrow 2(\ell +w)=26\Rightarrow \ell +w=13$. Area $40\Rightarrow \ell w=40$. Why this step? "Sum" and "product" of the two sides are exactly the two Vieta quantities — the problem is secretly a quadratic.
2. Build the quadratic whose roots are the sides: $x^2-(\ell +w)x+\ell w=0\Rightarrow x^2-13x+40=0.$ Why this step? Reverse-Vieta: given sum and product, this monic form reproduces them.
3. Solve. $x^2-13x+40=(x-5)(x-8)=0\Rightarrow x=5,8.$

Verify: sides $5$ and $8$. Perimeter $2(5+8)=26$ ✓; area $5\cdot 8=40$ ✓; both positive as lengths must be. (If the product had been negative we'd know the problem was impossible — a real rectangle can't have a negative-length side.)

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Cell I — Exam twist (find a hidden coefficient)

Forecast: you know the product must be $18$. If one root is $3$, what is the other — before touching $k$?

1. Use the product first. $\alpha \beta =\frac{c}{a}=\frac{18}{1}=18$. With $\alpha =3$: $3\beta =18\Rightarrow \beta =6.$ Why this step? The product ties both roots together, and the constant term is given, so it pins the second root immediately — no need for $k$ yet.
2. Now the sum gives $k$. $\alpha +\beta =3+6=9=-\frac{k}{1}\Rightarrow k=-9.$ Why this step? With both roots known, the sum rule turns into an equation for the unknown coefficient.

Verify: substitute $k=-9$: $x^2-9x+18=(x-3)(x-6)=0$, roots $3$ and $6$ ✓. Given root $3$ is present, and $6$ is the partner.

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Recall Which cell was hardest — quick self-quiz

Product is negative — what do the roots' signs do? ::: They have opposite signs. $c=0$ in $a{x}^2+bx+c=0$ — what is one root guaranteed to be? ::: Zero (product is $0$). Do Vieta's formulas still work when the roots are imaginary? ::: Yes — sum and product stay real for real coefficients. Given perimeter and area of a rectangle, why is Vieta's the fast route? ::: The two sides are roots of $x^2-(\text{sum})x+(\text{product})=0$.

See also: Polynomial Roots and Coefficients, Completing the Square, Systems of Equations, Descartes' Rule of Signs.