When you expand ( x − r 1 ) ( x − r 2 ) = x 2 − ( r 1 + r 2 ) x + r 1 r 2 (x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1 r_2 ( x − r 1 ) ( x − r 2 ) = x 2 − ( r 1 + r 2 ) x + r 1 r 2 coefficients directly encode the sum and product of the roots . Vieta's formulas let you extract root relationships WITHOUT solving the equation. They're reverse-engineering the expansion process.
Definition Vieta's FormulasFor a quadratic equation a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 α \alpha α β \beta β α + β = − b a \alpha + \beta = -\frac{b}{a} α + β = − a b α β = c a \alpha \beta = \frac{c}{a} α β = a c
For higher-degree polynomials, the pattern extends: coefficients are elementary symmetric polynomials in the roots.
Start with the factor form : If α \alpha α β \beta β a ( x − α ) ( x − β ) = 0 a(x - \alpha)(x - \beta) = 0 a ( x − α ) ( x − β ) = 0
Expand step-by-step :
a ( x − α ) ( x − β ) = a [ x 2 − β x − α x + α β ] a(x - \alpha)(x - \beta) = a[x^2 - \beta x - \alpha x + \alpha\beta] 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 - (\alpha + \beta)x + \alpha\beta] = a [ x 2 − ( α + β ) x + α β ] = a x 2 − a ( α + β ) x + a α β = ax^2 - a(\alpha + \beta)x + a\alpha\beta = a x 2 − a ( α + β ) x + a α β
Match with standard form a x 2 + b x + c ax^2 + bx + c a x 2 + b x + c
Coefficient of x x x b = − a ( α + β ) b = -a(\alpha + \beta) b = − a ( α + β ) α + β = − b a \alpha + \beta = -\frac{b}{a} α + β = − a b
Constant term: c = a α β c = a\alpha\beta c = a α β α β = c a \alpha\beta = \frac{c}{a} α β = a c
Why the negative sign in sum? Because we factor out ( x − α ) (x - \alpha) ( x − α )
General quadratic : a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 α + β = − b a , α β = c a \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} α + β = − a b , α β = a c
Cubic a x 3 + b x 2 + c x + d = 0 ax^3 + bx^2 + cx + d = 0 a x 3 + b x 2 + c x + d = 0 r 1 , r 2 , r 3 r_1, r_2, r_3 r 1 , r 2 , r 3 r 1 + r 2 + r 3 = − b a r_1 + r_2 + r_3 = -\frac{b}{a} r 1 + r 2 + r 3 = − a b r 1 r 2 + r 1 r 3 + r 2 r 3 = c a r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{c}{a} r 1 r 2 + r 1 r 3 + r 2 r 3 = a c r 1 r 2 r 3 = − d a r_1 r_2 r_3 = -\frac{d}{a} r 1 r 2 r 3 = − a d
Worked example Example 1: Finding Sum & Product Without SolvingProblem : For 3 x 2 − 12 x + 9 = 0 3x^2 - 12x + 9 = 0 3 x 2 − 12 x + 9 = 0 α + β \alpha + \beta α + β α β \alpha\beta α β
Solution :
Here a = 3 a = 3 a = 3 b = − 12 b = -12 b = − 12 c = 9 c = 9 c = 9
Step 1 : Apply Vieta's sum formula
α + β = − b a = − ( − 12 ) 3 = 12 3 = 4 \alpha + \beta = -\frac{b}{a} = -\frac{(-12)}{3} = \frac{12}{3} = 4 α + β = − a b = − 3 ( − 12 ) = 3 12 = 4
Why this step? The coefficient b = − 12 b = -12 b = − 12 − b = 12 -b = 12 − b = 12 a a a
Step 2 : Apply Vieta's product formula
α β = c a = 9 3 = 3 \alpha\beta = \frac{c}{a} = \frac{9}{3} = 3 α β = a c = 3 9 = 3
Why this step? The constant term relative to leading coefficient gives the product directly.
Answer : α + β = 4 \alpha + \beta = 4 α + β = 4 α β = 3 \alpha\beta = 3 α β = 3
Solution :
Step 1 : Calculate sum and product
α + β = 5 + ( − 2 ) = 3 \alpha + \beta = 5 + (-2) = 3 α + β = 5 + ( − 2 ) = 3 α β = 5 × ( − 2 ) = − 10 \alpha\beta = 5 \times (-2) = -10 α β = 5 × ( − 2 ) = − 10
Why? We need these to build the coefficients.
Step 2 : Use monic form x 2 − ( sum ) x + ( product ) = 0 x^2 - (\text{sum})x + (\text{product}) = 0 x 2 − ( sum ) x + ( product ) = 0 x 2 − 3 x + ( − 10 ) = 0 x^2 - 3x + (-10) = 0 x 2 − 3 x + ( − 10 ) = 0 x 2 − 3 x − 10 = 0 x^2 - 3x - 10 = 0 x 2 − 3 x − 10 = 0
Why monic form? When a = 1 a = 1 a = 1 p = − ( α + β ) p = -(\alpha + \beta) p = − ( α + β ) q = α β q = \alpha\beta q = α β
Step 3 : Verify by expansion
( x − 5 ) ( x + 2 ) = x 2 + 2 x − 5 x − 10 = x 2 − 3 x − 10 (x - 5)(x + 2) = x^2 + 2x - 5x - 10 = x^2 - 3x - 10 ( x − 5 ) ( x + 2 ) = x 2 + 2 x − 5 x − 10 = x 2 − 3 x − 10
Worked example Example 3: Findingα 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2 Problem : If 2 x 2 + 5 x − 3 = 0 2x^2 + 5x - 3 = 0 2 x 2 + 5 x − 3 = 0 α , β \alpha, \beta α , β α 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2
Solution :
Step 1 : Find sum and product from Vieta's
α + β = − 5 2 , α β = − 3 2 \alpha + \beta = -\frac{5}{2}, \quad \alpha\beta = \frac{-3}{2} α + β = − 2 5 , α β = 2 − 3
Step 2 : Use algebraic identity
α 2 + β 2 = ( α + β ) 2 − 2 α β \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta α 2 + β 2 = ( α + β ) 2 − 2 α β
Why this identity? Expanding ( α + β ) 2 = α 2 + 2 α β + β 2 (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 ( α + β ) 2 = α 2 + 2 α β + β 2 2 α β 2\alpha\beta 2 α β α 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2
Step 3 : Substitute
α 2 + β 2 = ( − 5 2 ) 2 − 2 ( − 3 2 ) \alpha^2 + \beta^2 = \left(-\frac{5}{2}\right)^2 - 2\left(-\frac{3}{2}\right) α 2 + β 2 = ( − 2 5 ) 2 − 2 ( − 2 3 ) = 25 4 + 6 2 = 25 4 + 12 4 = 37 4 = \frac{25}{4} + \frac{6}{2} = \frac{25}{4} + \frac{12}{4} = \frac{37}{4} = 4 25 + 2 6 = 4 25 + 4 12 = 4 37
Answer : 37 4 \frac{37}{4} 4 37
Worked example Example 4: Forming Equation with Transformed RootsProblem : If α , β \alpha, \beta α , β x 2 − 5 x + 3 = 0 x^2 - 5x + 3 = 0 x 2 − 5 x + 3 = 0 α 2 , β 2 \alpha^2, \beta^2 α 2 , β 2
Solution :
Step 1 : Original root properties
α + β = 5 , α β = 3 \alpha + \beta = 5, \quad \alpha\beta = 3 α + β = 5 , α β = 3
Step 2 : Find sum of new roots
α 2 + β 2 = ( α + β ) 2 − 2 α β = 25 − 6 = 19 \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 = 19 α 2 + β 2 = ( α + β ) 2 − 2 α β = 25 − 6 = 19
Step 3 : Find product of new roots
α 2 β 2 = ( α β ) 2 = 3 2 = 9 \alpha^2 \beta^2 = (\alpha\beta)^2 = 3^2 = 9 α 2 β 2 = ( α β ) 2 = 3 2 = 9
Why? The product of squares is the square of the product.
Step 4 : Construct new equation
x 2 − ( α 2 + β 2 ) x + α 2 β 2 = 0 x^2 -(\alpha^2 + \beta^2)x + \alpha^2\beta^2 = 0 x 2 − ( α 2 + β 2 ) x + α 2 β 2 = 0 x 2 − 19 x + 9 = 0 x^2 - 19x + 9 = 0 x 2 − 19 x + 9 = 0
Common mistake Mistake 1: Forgetting the Negative Sign in SumWrong thinking : "The coefficient of x x x b b b b a \frac{b}{a} a b
Why it feels right : We see b b b
The fix : The sum is − b a -\frac{b}{a} − a b ( x − α ) ( x − β ) (x - \alpha)(x - \beta) ( x − α ) ( x − β ) − ( α + β ) -(\alpha + \beta) − ( α + β )
Steel-man : The confusion comes from x 2 + p x + q x^2 + px + q x 2 + p x + q α + β = − p \alpha + \beta = -p α + β = − p
Common mistake Mistake 2: Wrong Formula for Non-Monic EquationsWrong thinking : For 3 x 2 + 6 x + 2 = 0 3x^2 + 6x + 2 = 0 3 x 2 + 6 x + 2 = 0 α + β = − 6 \alpha + \beta = -6 α + β = − 6
Why it feels right : Monic form has α + β = − p \alpha + \beta = -p α + β = − p x 2 + p x + q = 0 x^2 + px + q = 0 x 2 + p x + q = 0
The fix : ALWAYS divide by a a a α + β = − 6 3 = − 2 \alpha + \beta = -\frac{6}{3} = -2 α + β = − 3 6 = − 2 − 6 -6 − 6
Key check : If you don't divide by a a a
Why it feels right : The square of a sum notation looks similar.
The fix :
( α + β ) 2 = α 2 + 2 α β + β 2 (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 ( α + β ) 2 = α 2 + 2 α β + β 2 α 2 + β 2 = ( α + β ) 2 − 2 α β \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta α 2 + β 2 = ( α + β ) 2 − 2 α β
Why this works : The middle term 2 α β 2\alpha\beta 2 α β
1. Checking factorization : If you claim ( x − 2 ) ( x − 7 ) = x 2 − 9 x + 14 (x - 2)(x - 7) = x^2 - 9x + 14 ( x − 2 ) ( x − 7 ) = x 2 − 9 x + 14 2 + 7 = 9 2 + 7 = 9 2 + 7 = 9 2 × 7 = 14 2 \times 7 = 14 2 × 7 = 14 − ( − 9 ) = 9 -(-9) = 9 − ( − 9 ) = 9 14 14 14
2. Sum/product constraints : If α + β = 4 \alpha + \beta = 4 α + β = 4 α β = − 5 \alpha\beta = -5 α β = − 5 x 2 − 4 x − 5 = 0 x^2 - 4x - 5 = 0 x 2 − 4 x − 5 = 0
3. Symmetric expressions : Any symmetric function of roots (like α 3 + β 3 \alpha^3 + \beta^3 α 3 + β 3 1 α + 1 β \frac{1}{\alpha} + \frac{1}{\beta} α 1 + β 1 α + β \alpha + \beta α + β α β \alpha\beta α β 1 α + 1 β = α + β α β \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} α 1 + β 1 = α β α + β
4. Root bounds : If α β > 0 \alpha\beta > 0 α β > 0 α + β > 0 \alpha + \beta > 0 α + β > 0
Mnemonic Remember the Signs"SNAP" — S um is N egative b b b a a a P roduct is P ositive c c c a a a c c c
Think: "Sum? Negate b. Product? Plain c. " (after dividing by a a a
Recall Explain Like You're Teaching a Younger StudentImagine you're making a secret code. You have two mystery numbers, let's call them α \alpha α β \beta β
Now, there's a magic box called a quadratic equation that looks like a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 a a a b b b c c c
Vieta's formulas are the decoder :
If you add your two numbers: α + β = − b a \alpha + \beta = -\frac{b}{a} α + β = − a b
If you multiply them: α × β = c a \alpha \times \beta = \frac{c}{a} α × β = a c
Why does this work? Because when mathematicians multiply ( x − α ) ( x − β ) (x - \alpha)(x - \beta) ( x − α ) ( x − β ) x 2 − ( α + β ) x + α β x^2 - (\alpha+\beta)x + \alpha\beta x 2 − ( α + β ) x + α β
Real use : You can know things about the roots WITHOUT actually solving! Like if someone asks "Do the roots add up to 7?" you just check − b a -\frac{b}{a} − a b
#flashcards/maths
For a quadratic a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 α , β \alpha, \beta α , β α + β \alpha + \beta α + β For a quadratic a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 α , β \alpha, \beta α , β α β \alpha\beta α β Why is there a negative sign in the sum of roots formula? Because the factor form is
( x − α ) ( x − β ) (x - \alpha)(x - \beta) ( x − α ) ( x − β ) , so when expanded the sum appears as
− ( α + β ) -(\alpha + \beta) − ( α + β ) .
How do you construct a quadratic with roots 3 and -7? Sum =
− 4 -4 − 4 , product =
− 21 -21 − 21 , so equation is
x 2 + 4 x − 21 = 0 x^2 + 4x - 21 = 0 x 2 + 4 x − 21 = 0 (using
x 2 − ( sum ) x + ( product ) = 0 x^2 - (\text{sum})x + (\text{product}) = 0 x 2 − ( sum ) x + ( product ) = 0 ).
If α + β = 5 \alpha + \beta = 5 α + β = 5 α β = 6 \alpha\beta = 6 α β = 6 α 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2 ( α + β ) 2 − 2 α β = 25 − 12 = 13 (\alpha + \beta)^2 - 2\alpha\beta = 25 - 12 = 13 ( α + β ) 2 − 2 α β = 25 − 12 = 13 .
For 3 x 2 − 9 x + 6 = 0 3x^2 - 9x + 6 = 0 3 x 2 − 9 x + 6 = 0 − − 9 3 = 3 -\frac{-9}{3} = 3 − 3 − 9 = 3 .
What is the formula for 1 α + 1 β \frac{1}{\alpha} + \frac{1}{\beta} α 1 + β 1 α + β α β \frac{\alpha + \beta}{\alpha\beta} α β α + β .
In a cubic a x 3 + b x 2 + c x + d = 0 ax^3 + bx^2 + cx + d = 0 a x 3 + b x 2 + c x + d = 0 r 1 + r 2 + r 3 r_1 + r_2 + r_3 r 1 + r 2 + r 3
Quadratic Formula — Vieta's gives relationships without explicit root valuesFactoring Quadratics — Reverse process: Vieta's → coefficients, factoring → rootsSymmetric Polynomials — Vieta's formulas are elementary symmetric functionsPolynomial Roots and Coefficients — Generalizes to higher degreesCompleting the Square — Alternative derivation pathDiscriminant — Δ = b 2 − 4 a c = ( α − β ) 2 ⋅ a 2 \Delta = b^2 - 4ac = (\alpha - \beta)^2 \cdot a^2 Δ = b 2 − 4 a c = ( α − β ) 2 ⋅ a 2 Systems of Equations — Use Vieta's to set up systems from root conditionsDescartes' Rule of Signs — Combined with Vieta's for root properties
Master Vieta's and you'll never waste time solving quadratics when you only need root relationships.
Factor form a x-alpha x-beta
a x^2 - a sum x + a product
Elementary symmetric polynomials
Example: find sum and product
Example: construct from roots
Intuition Hinglish mein samjhoVieta's formulas ek bahut powerful shortcut hai jab apko quadratic equation ke roots ka sum aur product chahiye, lekin exact values nikalne ki zarurat nahi. Dekho, jab aap ( x − α ) ( x − β ) (x - \alpha)(x - \beta) ( x − α ) ( x − β ) a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 − b / a -b/a − b / a c / a c/a c / a
Yeh kyu important hai? Kyunki bahut sare problems mein tumhe sirf relationships chahiye, exact roots nahi. Jaise agar koi puche "roots ka square ka sum kya hai?" toh tum directly identity use kar sakte ho: α 2 + β 2 = ( α + β ) 2 − 2 α β \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta α 2 + β 2 = ( α + β ) 2 − 2 α β
Ek common mistake jo students karte hain: woh negative sign bhul jate hain sum mein. Yad rakho, sum hamesha − b / a -b/a − b / a b / a b/a b / a ( x − α ) (x - \alpha) ( x − α ) a ≠ 1 a \neq 1 a = 1 a a a
Vieta's ka sabse practical use? New equations banana jab transformed roots diye ho. Agar original equation ke roots α , β \alpha, \beta α , β α 2 , β 2 \alpha^2, \beta^2 α 2 , β 2
