Vieta's formulas — sum and product of roots

2.1.19 · Maths › Algebra — Introduction & Intermediate

What Are Vieta's Formulas?

Higher-degree polynomials ke liye bhi yahi pattern extend hota hai: coefficients roots mein elementary symmetric polynomials hote hain.

Why Does This Work? — Derivation from Scratch

Factor form se shuru karo: Agar $α$ aur $β$ roots hain, toh: $a (x - α) (x - β) = 0$

Step-by-step expand karo: $a (x - α) (x - β) = a [x^{2} - β x - α x + α β]$ $= a [x^{2} - (α + β) x + α β]$ $= a x^{2} - a (α + β) x + a α β$

Standard form $a x^{2} + b x + c$ se match karo:

$x$ ka coefficient: $b = - a (α + β)$ → $α + β = - \frac{b}{a}$
Constant term: $c = a α β$ → $α β = \frac{c}{a}$

Sum mein negative sign kyun? Kyunki hum $(x - α)$ factor karte hain (minus roots), toh jab expand karte hain toh sum opposite sign ke saath aata hai.

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

General quadratic: $a x^{2} + b x + c = 0$ $α + β = - \frac{b}{a}, α β = \frac{c}{a}$

Cubic $a x^{3} + b x^{2} + c x + d = 0$ jiske roots $r_{1}, r_{2}, r_{3}$ hain: $r_{1} + r_{2} + r_{3} = - \frac{b}{a}$ $r_{1} r_{2} + r_{1} r_{3} + r_{2} r_{3} = \frac{c}{a}$ $r_{1} r_{2} r_{3} = - \frac{d}{a}$

Worked Examples — Learn by Doing

Solution: Yahan $a = 3$ , $b = - 12$ , $c = 9$ hai.

Step 1: Vieta's sum formula apply karo $α + β = - \frac{b}{a} = - \frac{( - 12 )}{3} = \frac{12}{3} = 4$

Ye step kyun? Coefficient $b = - 12$ negative hai, toh $- b = 12$ positive hai. $a$ se divide karna normalize karta hai.

Step 2: Vieta's product formula apply karo $α β = \frac{c}{a} = \frac{9}{3} = 3$

Ye step kyun? Constant term ko leading coefficient se divide karo toh seedha product milta hai.

Answer: $α + β = 4$ , $α β = 3$

Solution: Step 1: Sum aur product calculate karo $α + β = 5 + (- 2) = 3$ $α β = 5 \times (- 2) = - 10$

Kyun? Humein coefficients banane ke liye inki zaroorat hai.

Step 2: Monic form $x^{2} - (sum) x + (product) = 0$ use karo $x^{2} - 3 x + (- 10) = 0$ $x^{2} - 3 x - 10 = 0$

Monic form kyun? Jab $a = 1$ hota hai, Vieta's hume seedha $p = - (α + β)$ aur $q = α β$ deta hai.

Step 3: Expansion se verify karo $(x - 5) (x + 2) = x^{2} + 2 x - 5 x - 10 = x^{2} - 3 x - 10$ ✓

Solution: Step 1: Vieta's se sum aur product nikalo $α + β = - \frac{5}{2}, α β = \frac{- 3}{2}$

Step 2: Algebraic identity use karo $α^{2} + β^{2} = (α + β)^{2} - 2 α β$

Ye identity kyun? $(α + β)^{2} = α^{2} + 2 α β + β^{2}$ expand karo, phir $α^{2} + β^{2}$ isolate karne ke liye $2 α β$ subtract karo.

Step 3: Substitute karo $α^{2} + β^{2} = (- \frac{5}{2})^{2} - 2 (- \frac{3}{2})$ $= \frac{25}{4} + \frac{6}{2} = \frac{25}{4} + \frac{12}{4} = \frac{37}{4}$

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

Solution: Step 1: Original root properties $α + β = 5, α β = 3$

Step 2: Naye roots ka sum nikalo $α^{2} + β^{2} = (α + β)^{2} - 2 α β = 25 - 6 = 19$

Step 3: Naye roots ka product nikalo $α^{2} β^{2} = (α β)^{2} = 3^{2} = 9$

Kyun? Squares ka product, product ka square hota hai.

Step 4: Nayi equation construct karo $x^{2} - (α^{2} + β^{2}) x + α^{2} β^{2} = 0$ $x^{2} - 19 x + 9 = 0$

Common Mistakes & Fixes

Ye sahi kyun lagta hai: Hum $b$ seedha equation mein dekhte hain.

Fix: Sum $- \frac{b}{a}$ hai kyunki factor form $(x - α) (x - β)$ hai. Jab expand karte ho, toh $- (α + β)$ coefficient ke roop mein aata hai.

Steel-man: Confusion $x^{2} + p x + q$ form se aata hai jahan hum $α + β = - p$ likhte hain. Sign convention formula mein already built-in hai.

Ye sahi kyun lagta hai: Monic form mein $α + β = - p$ hota hai jab equation $x^{2} + p x + q = 0$ ho.

Fix: HAMESHA pehle $a$ se divide karo. Yahan $α + β = - \frac{6}{3} = - 2$ hai, $- 6$ nahi.

Key check: Agar $a$ se divide nahi kiya, toh tum ek alag equation ke liye relationships nikal rahe ho.

Ye sahi kyun lagta hai: Sum ka square notation milta-julta lagta hai.

Fix: $(α + β)^{2} = α^{2} + 2 α β + β^{2}$ Toh $α^{2} + β^{2} = (α + β)^{2} - 2 α β$

Ye kyun kaam karta hai: Middle term $2 α β$ ko subtract karna padta hai.

Power Techniques — 80/20 Applications

1. Factorization check karna: Agar tum claim karte ho ki $(x - 2) (x - 7) = x^{2} - 9 x + 14$ , toh verify karo: sum $2 + 7 = 9$ hona chahiye, product $2 \times 7 = 14$ . Check karo $- (- 9) = 9$ ✓ aur $14$ ✓.

2. Sum/product constraints: Agar $α + β = 4$ aur $α β = - 5$ , toh equation hai $x^{2} - 4 x - 5 = 0$ . Instant equation construction.

3. Symmetric expressions: Roots ka koi bhi symmetric function (jaise $α^{3} + β^{3}$ , $\frac{1}{α} + \frac{1}{β}$ ) ko $α + β$ aur $α β$ ke through express kiya ja sakta hai: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$

4. Root bounds: Agar $α β > 0$ , toh roots same sign ke hain. Agar $α + β > 0$ bhi hai, toh dono positive hain.

Socho: "Sum? Negate b. Product? Plain c." ( $a$ se divide karne ke baad).

Feynman Technique — Explain to a 12-Year-Old

Recall Aisa Explain Karo Jaise Kisi Chote Student Ko Padha Rahe Ho

Socho tum ek secret code bana rahe ho. Tumhare paas do mystery numbers hain, chaliye unhe $α$ aur $β$ bolte hain. Tum kisi ko nahi batate wo kya hain, lekin tum unhe ZAROOR batate ho ki unhe jodne par kya milta hai aur gunna karne par kya milta hai.

Ab, ek magic box hai jise quadratic equation kehte hain jo $a x^{2} + b x + c = 0$ jaisa dikhta hai. Trick yeh hai: wo letters $a$ , $b$ , $c$ mein tumhare mystery numbers ke baare mein chuppe clues hain!

Vieta's formulas wo decoder hain:

Agar tum apne do numbers jodo: $α + β = - \frac{b}{a}$ (beech wala part, ulta)
Agar tum unhe gunna karo: $α \times β = \frac{c}{a}$ (aakhiri part)

Ye kyun kaam karta hai? Kyunki jab mathematicians $(x - α) (x - β)$ multiply karte hain, unhe exactly $x^{2} - (α + β) x + α β$ milta hai. Toh equation khud hi expanded form hai, aur Vieta's humein backwards kaam karne mein help karta hai.

Real use: Tum roots ke baare mein kuch jaan sakte ho bina actually solve kiye! Jaise agar koi poochhe "Kya roots ka sum 7 hai?" tum bas $- \frac{b}{a}$ check karo. Quadratic formula ki zaroorat nahi!

Active Recall Flashcards

#flashcards/maths

For a quadratic

a x^{2} + b x + c = 0

with roots

α, β

, what is

α + β

- \frac{b}{a}

For a quadratic

a x^{2} + b x + c = 0

with roots

α, β

, what is

α β

\frac{c}{a}

Sum of roots formula mein negative sign kyun hota hai?

Kyunki factor form

(x - α) (x - β)

hai, toh expand karne par sum

- (α + β)

ke roop mein aata hai.

3 aur -7 roots wali quadratic kaise construct karoge?

Sum =

- 4

, product =

- 21

, toh equation hai

x^{2} + 4 x - 21 = 0

(

x^{2} - (sum) x + (product) = 0

use karke).

Agar

α + β = 5

aur

α β = 6

, toh

α^{2} + β^{2}

kya hai?

(α + β)^{2} - 2 α β = 25 - 12 = 13

3 x^{2} - 9 x + 6 = 0

ke liye roots ka sum kya hai?

- \frac{- 9}{3} = 3

Vieta's ke terms mein

\frac{1}{α} + \frac{1}{β}

ka formula kya hai?

\frac{α + β}{α β}

Ek cubic

a x^{3} + b x^{2} + c x + d = 0

mein

r_{1} + r_{2} + r_{3}

kya hai?

- \frac{b}{a}

Connections

Quadratic Formula — Vieta's explicit root values ke bina relationships deta hai
Factoring Quadratics — Reverse process: Vieta's → coefficients, factoring → roots
Symmetric Polynomials — Vieta's formulas elementary symmetric functions hain
Polynomial Roots and Coefficients — Higher degrees tak generalize hota hai
Completing the Square — Alternative derivation path
Discriminant — $Δ = b^{2} - 4 a c = (α - β)^{2} \cdot a^{2}$ root spacing se related hai
Systems of Equations — Root conditions se systems set up karne ke liye Vieta's use karo
Descartes' Rule of Signs — Root properties ke liye Vieta's ke saath combine karo

Vieta's master karo aur tum kabhi quadratics solve karne mein time waste nahi karoge jab sirf root relationships chahiye hoti hain.