Vieta's formulas — sum and product of roots: Exercises

Before we start, let me re-anchor the two facts we lean on the entire way down. For a quadratic written $a x^{2} + b x + c = 0$ — that is, $a$ is the number in front of $x^{2}$ , $b$ the number in front of $x$ , and $c$ the lonely constant — the two roots $α$ (alpha) and $β$ (beta) satisfy:

Keep the picture below in your head the whole time: the coefficients are just the expanded shadow of the factored form $a (x - α) (x - β)$ .

Figure — Vieta's formulas — sum and product of roots

Related tools you may reach for: Quadratic Formula, Factoring Quadratics, Completing the Square, Discriminant, Symmetric Polynomials, Polynomial Roots and Coefficients, Systems of Equations, Descartes' Rule of Signs.

Level 1 — Recognition

Goal: read the coefficients and quote the right formula.

Recall Solution L1-a

Identify the coefficients. Here $a = 1$ , $b = - 7$ , $c = 12$ . Sum: $α + β = - \frac{b}{a} = - \frac{- 7}{1} = 7$ . Product: $α β = \frac{c}{a} = \frac{12}{1} = 12$ . (Sanity: this factors as $(x - 3) (x - 4)$ , and indeed $3 + 4 = 7$ , $3 \cdot 4 = 12$ .) Answer: $α + β = 7$ , $α β = 12$ .

Recall Solution L1-b

$a = 1$ , $b = 9$ , $c = 20$ . Sum: $- \frac{9}{1} = - 9$ . Product: $\frac{20}{1} = 20$ . (Factors as $(x + 4) (x + 5)$ : $- 4 + - 5 = - 9$ , $(- 4) (- 5) = 20$ .) Answer: $α + β = - 9$ , $α β = 20$ .

Level 2 — Application

Goal: handle a non-monic $a$ and negative coefficients, then run the formula.

Recall Solution L2-a

$a = 2$ , $b = - 3$ , $c = - 5$ . Sum: $- \frac{b}{a} = - \frac{- 3}{2} = \frac{3}{2}$ . Product: $\frac{c}{a} = \frac{- 5}{2} = - \frac{5}{2}$ . Answer: $α + β = \frac{3}{2}$ , $α β = - \frac{5}{2}$ .

Recall Solution L2-b

Use the construction rule: a monic quadratic with roots $α, β$ is $x^{2} - (α + β) x + α β = 0.$ Sum: $6 + (- \frac{1}{2}) = \frac{11}{2}$ . Product: $6 \times (- \frac{1}{2}) = - 3$ . Plug in: $x^{2} - \frac{11}{2} x - 3 = 0.$ Multiply through by $2$ to clear the fraction (optional, keeps the same roots): $2 x^{2} - 11 x - 6 = 0.$ (Check: $(x - 6) (2 x + 1) = 2 x^{2} + x - 12 x - 6 = 2 x^{2} - 11 x - 6$ . ✓) Answer: $x^{2} - \frac{11}{2} x - 3 = 0$ , or equivalently $2 x^{2} - 11 x - 6 = 0$ .

Recall Solution L2-c

Why not solve for the roots? Because $\frac{1}{α} + \frac{1}{β}$ is a symmetric expression — it doesn't care which root is which — so it can be written entirely from the sum and product. Combine the fractions: $\frac{1}{α} + \frac{1}{β} = \frac{β + α}{α β} = \frac{α + β}{α β} .$ From Vieta: $α + β = - \frac{5}{2}$ , $α β = - \frac{3}{2}$ . $\frac{α + β}{α β} = \frac{- 5/2}{- 3/2} = \frac{- 5}{- 3} = \frac{5}{3} .$ Answer: $\frac{5}{3}$ .

Level 3 — Analysis

Goal: combine Vieta with algebraic identities to reach expressions the roots satisfy.

Recall Solution L3-a

Vieta: $α + β = - \frac{- 12}{3} = 4$ , $α β = \frac{4}{3}$ . Identity we need: squaring a sum gives a cross term, so $(α + β)^{2} = α^{2} + 2 α β + β^{2} \Rightarrow α^{2} + β^{2} = (α + β)^{2} - 2 α β .$ Substitute: $α^{2} + β^{2} = 4^{2} - 2 \cdot \frac{4}{3} = 16 - \frac{8}{3} = \frac{48}{3} - \frac{8}{3} = \frac{40}{3} .$ Answer: $\frac{40}{3}$ .

Recall Solution L3-b

Identity: $(α - β)^{2} = α^{2} - 2 α β + β^{2} = (α + β)^{2} - 4 α β$ . (Why this form? It uses only the two Vieta quantities — no need to know the roots individually, and $(α - β)^{2}$ is symmetric even though $α - β$ alone is not.) $(α - β)^{2} = 4^{2} - 4 \cdot \frac{4}{3} = 16 - \frac{16}{3} = \frac{48 - 16}{3} = \frac{32}{3} .$ Answer: $\frac{32}{3}$ . (Aside: $(α - β)^{2} = \frac{b ^{2} - 4 a c}{a ^{2}}$ — the Discriminant over $a^{2}$ . Here $\frac{144 - 48}{9} = \frac{96}{9} = \frac{32}{3}$ ✓.)

Recall Solution L3-c

Vieta: $α + β = 5$ , $α β = 6$ . Identity for a sum of cubes (the one that stays symmetric): $α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β) .$ (Where from? $(α + β)^{3} = α^{3} + 3 α^{2} β + 3 α β^{2} + β^{3} = α^{3} + β^{3} + 3 α β (α + β)$ ; rearrange.) $α^{3} + β^{3} = 5^{3} - 3 \cdot 6 \cdot 5 = 125 - 90 = 35.$ Answer: $35$ .

Level 4 — Synthesis

Goal: build a brand-new equation whose roots are transformed versions of the old ones.

Recall Solution L4-a

Old Vieta: $α + β = 6$ , $α β = 4$ . To build the new equation $x^{2} - (new sum) x + (new product) = 0$ , I need the sum and product of the new roots $α^{2}, β^{2}$ . New sum: $α^{2} + β^{2} = (α + β)^{2} - 2 α β = 36 - 8 = 28.$ New product: $α^{2} β^{2} = (α β)^{2} = 4^{2} = 16.$ Assemble: $x^{2} - 28 x + 16 = 0.$ Answer: $x^{2} - 28 x + 16 = 0$ .

Recall Solution L4-b

Old Vieta: $α + β = \frac{7}{2}$ , $α β = \frac{3}{2}$ . New sum $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{7/2}{3/2} = \frac{7}{3} .$ New product $\frac{1}{α} \cdot \frac{1}{β} = \frac{1}{α β} = \frac{1}{3/2} = \frac{2}{3} .$ Assemble monic: $x^{2} - \frac{7}{3} x + \frac{2}{3} = 0 ⟹ 3 x^{2} - 7 x + 2 = 0.$ (Neat shortcut: reversing the coefficients of $2 x^{2} - 7 x + 3$ gives $3 x^{2} - 7 x + 2$ — reciprocal roots always reverse the coefficient list. ✓) Answer: $3 x^{2} - 7 x + 2 = 0$ .

Recall Solution L4-c

Old Vieta: $α + β = - 4$ , $α β = 1$ . New sum: $(α + 2) + (β + 2) = (α + β) + 4 = - 4 + 4 = 0.$ New product: $(α + 2) (β + 2) = α β + 2 (α + β) + 4 = 1 + 2 (- 4) + 4 = 1 - 8 + 4 = - 3.$ Assemble: $x^{2} - (0) x + (- 3) = 0 ⟹ x^{2} - 3 = 0.$ Answer: $x^{2} - 3 = 0$ .

Level 5 — Mastery

Goal: degenerate and boundary cases — repeated roots, complex roots, sign reasoning, and a reverse problem.

Recall Solution L5-a

"Repeated root" means $α = β$ . Call the common value $r$ . Vieta: sum $= 2 r = k$ and product $= r^{2} = 9$ . From the product, $r = \pm 3$ .

If $r = 3$ : $k = 2 (3) = 6$ .
If $r = - 3$ : $k = 2 (- 3) = - 6$ . (Cross-check with the Discriminant: equal roots need $b^{2} - 4 a c = 0$ , i.e. $k^{2} - 36 = 0 \Rightarrow k = \pm 6$ . ✓) Answer: $k = 6$ (root $3$ ) or $k = - 6$ (root $- 3$ ).

Recall Solution L5-b

Vieta's formulas hold whether roots are real or complex — they come from expansion, not from the Quadratic Formula. So: Vieta: $α + β = - 1$ , $α β = 1$ . $α^{2} + β^{2} = (α + β)^{2} - 2 α β = (- 1)^{2} - 2 (1) = 1 - 2 = - 1.$ Answer: $- 1$ . (A negative sum of squares is fine — the roots aren't real numbers, so their squares needn't be positive.)

Recall Solution L5-c

Vieta: $α + β = 7 > 0$ , $α β = 10 > 0$ . Reasoning like Descartes' Rule of Signs but through Vieta:

Product $> 0$ ⟹ the roots have the same sign (a positive product can't come from one positive and one negative).
Same sign and sum $> 0$ ⟹ both must be positive. Answer: both roots positive. (Indeed $x^{2} - 7 x + 10 = (x - 2) (x - 5)$ , roots $2, 5$ . ✓)

Recall Solution L5-d

This is a small Systems of Equations problem, but Vieta turns it into one quadratic. If the numbers are the roots, then $x^{2} - (sum) x + (product) = x^{2} - 5 x - 14 = 0.$ Factor (or use the Quadratic Formula): $(x - 7) (x + 2) = 0$ , so $x = 7$ or $x = - 2$ . Answer: the numbers are $7$ and $- 2$ . (Check: $7 + (- 2) = 5$ , $7 \times (- 2) = - 14$ . ✓)

Wrap-up recall

Recall One-line reminders

Sum of roots ::: $α + β = - \frac{b}{a}$ Product of roots ::: $α β = \frac{c}{a}$ Build monic quadratic from roots ::: $x^{2} - (α + β) x + α β = 0$ New product when roots are squared ::: $(α β)^{2}$ (square it, don't double it) Does Vieta need real roots? ::: No — it comes from expansion, works for complex roots too