If you know where a parabola touches the x-axis (its roots), you can reverse-engineer the entire equation. Think of roots as "DNA" — they uniquely determine the quadratic's structure. When x = α x = \alpha x = α ( x − α ) (x - \alpha) ( x − α ) α \alpha α β \beta β ( x − α ) ( x − β ) (x - \alpha)(x - \beta) ( x − α ) ( x − β )
A quadratic equation has roots α \alpha α β \beta β
Step 1: What does "root" mean?
If α \alpha α P ( x ) P(x) P ( x ) P ( α ) = 0 P(\alpha) = 0 P ( α ) = 0 ( x − α ) (x - \alpha) ( x − α )
Step 2: Two roots means two factors
If both α \alpha α β \beta β
( x − α ) (x - \alpha) ( x − α ) P ( α ) = 0 P(\alpha) = 0 P ( α ) = 0 ( x − β ) (x - \beta) ( x − β ) P ( β ) = 0 P(\beta) = 0 P ( β ) = 0
Step 3: Multiply the factors
The simplest polynomial with both factors is their product:
Why the k k k Roots only tell us where the parabola crosses the x-axis, not how stretched it is. All these parabolas have the same roots but different "widths":
k = 1 k = 1 k = 1 k = 2 k = 2 k = 2 k = 0.5 k = 0.5 k = 0.5
Notice the coefficients! If a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0
Definition Vieta's Formulas (Derived)From k ( x − α ) ( x − β ) = k x 2 − k ( α + β ) x + k α β k(x - \alpha)(x - \beta) = kx^2 - k(\alpha+\beta)x + k\alpha\beta k ( x − α ) ( x − β ) = k x 2 − k ( α + β ) x + k α β
Sum of roots: α + β = − b a \alpha + \beta = -\frac{b}{a} α + β = − a b x x x
Product of roots: α β = c a \alpha \beta = \frac{c}{a} α β = a c
Why this matters: You can form a quadratic from just sum and product, without knowing individual roots!
x 2 − ( sum of roots ) x + ( product of roots ) = 0 x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 x 2 − ( sum of roots ) x + ( product of roots ) = 0
Worked example Example 1: Given Individual RootsProblem: Form a quadratic equation with roots α = 3 \alpha = 3 α = 3 β = − 5 \beta = -5 β = − 5 Solution:
Step 1: Use the factor form
P ( x ) = k ( x − 3 ) ( x − ( − 5 ) ) = k ( x − 3 ) ( x + 5 ) P(x) = k(x - 3)(x - (-5)) = k(x - 3)(x + 5) P ( x ) = k ( x − 3 ) ( x − ( − 5 )) = k ( x − 3 ) ( x + 5 )
Why this step? Each root contributes a factor that zeroes out at that root.
Step 2: Expand (taking k = 1 k = 1 k = 1 P ( x ) = ( x − 3 ) ( x + 5 ) P(x) = (x - 3)(x + 5) P ( x ) = ( x − 3 ) ( x + 5 ) = x 2 + 5 x − 3 x − 15 = x^2 + 5x - 3x - 15 = x 2 + 5 x − 3 x − 15 = x 2 + 2 x − 15 = x^2 + 2x - 15 = x 2 + 2 x − 15 Why expand? Standard form a x 2 + b x + c ax^2 + bx + c a x 2 + b x + c
Answer: x 2 + 2 x − 15 = 0 x^2 + 2x - 15 = 0 x 2 + 2 x − 15 = 0
Verify: Sum = 3 + ( − 5 ) = − 2 = − 2 1 = 3 + (-5) = -2= -\frac{2}{1} = 3 + ( − 5 ) = − 2 = − 1 2 = 3 × ( − 5 ) = − 15 = − 15 1 = 3 \times (-5) = -15 = \frac{-15}{1} = 3 × ( − 5 ) = − 15 = 1 − 15
Worked example Example 2: Given Sum and ProductProblem: Form a quadratic whose roots have sum = 7 = 7 = 7 = 10 = 10 = 10
Solution:
Direct formula:
x 2 − ( sum ) x + ( product ) = 0 x^2 - (\text{sum})x + (\text{product}) = 0 x 2 − ( sum ) x + ( product ) = 0 x 2 − 7 x + 10 = 0 x^2 - 7x + 10 = 0 x 2 − 7 x + 10 = 0 Why this works? From Vieta's formulas: α + β = 7 \alpha + \beta = 7 α + β = 7 α β = 10 \alpha\beta = 10 α β = 10 ( x − α ) ( x − β ) (x - \alpha)(x - \beta) ( x − α ) ( x − β )
Finding actual roots (bonus check):
Factor: ( x − 5 ) ( x − 2 ) = 0 (x - 5)(x - 2) = 0 ( x − 5 ) ( x − 2 ) = 0 α = 5 , β = 2 \alpha = 5, \beta = 2 α = 5 , β = 2 5 + 2 = 7 5 + 2 = 7 5 + 2 = 7 5 × 2 = 10 5 \times 2 = 10 5 × 2 = 10
Worked example Example 3: Roots with RadicalsProblem: Roots are 2 + 3 2+ \sqrt{3} 2 + 3 2 − 3 2 - \sqrt{3} 2 − 3
Solution:
Method 1 (factor form):
P ( x ) = [ x − ( 2 + 3 ) ] [ x − ( 2 − 3 ) ] P(x) = [x - (2 + \sqrt{3})][x - (2 - \sqrt{3})] P ( x ) = [ x − ( 2 + 3 )] [ x − ( 2 − 3 )] Expand using difference of squares pattern:
Let a = x − 2 a = x - 2 a = x − 2 P ( x ) = ( a − 3 ) ( a + 3 ) = a 2 − 3 P(x) = (a - \sqrt{3})(a + \sqrt{3}) = a^2 - 3 P ( x ) = ( a − 3 ) ( a + 3 ) = a 2 − 3 = ( x − 2 ) 2 − 3 = (x - 2)^2 - 3 = ( x − 2 ) 2 − 3 = x 2 − 4 x + 4 − 3 = x^2 - 4x + 4 - 3 = x 2 − 4 x + 4 − 3 = x 2 − 4 x + 1 = x^2 - 4x + 1 = x 2 − 4 x + 1
Why this step? Recognizing conjugate pairs ( a − b ) ( a + b ) = a 2 − b 2 (a - b)(a + b) = a^2 - b^2 ( a − b ) ( a + b ) = a 2 − b 2
Method 2 (sum and product):
Sum: ( 2 + 3 ) + ( 2 − 3 ) = 4 (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 ( 2 + 3 ) + ( 2 − 3 ) = 4
Product: ( 2 + 3 ) ( 2 − 3 ) = 4 − 3 = 1 (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 ( 2 + 3 ) ( 2 − 3 ) = 4 − 3 = 1
Equation: x 2 − 4 x + 1 = 0 x^2 - 4x + 1 = 0 x 2 − 4 x + 1 = 0
Answer: x 2 − 4 x + 1 = 0 x^2 - 4x + 1 = 0 x 2 − 4 x + 1 = 0
Worked example Example 4: Scaled QuadraticProblem: Form a quadratic with roots − 2 -2 − 2 3 3 3 ( 1 , 8 ) (1, 8) ( 1 , 8 )
Solution:
Step 1: Generic form with roots
P ( x ) = k ( x + 2 ) ( x − 3 ) P(x) = k(x + 2)(x - 3) P ( x ) = k ( x + 2 ) ( x − 3 ) Step 2: Use the point condition P ( 1 ) = 8 P(1) = 8 P ( 1 ) = 8 8 = k ( 1 + 2 ) ( 1 − 3 ) 8 = k(1 + 2)(1 - 3) 8 = k ( 1 + 2 ) ( 1 − 3 ) 8 = k ( 3 ) ( − 2 ) 8 = k(3)(-2) 8 = k ( 3 ) ( − 2 ) 8 = − 6 k 8 = -6k 8 = − 6 k k = − 4 3 k = -\frac{4}{3} k = − 3 4
Why this step? The value of k k k k k k
Step 3: Write final equation
P ( x ) = − 4 3 ( x + 2 ) ( x − 3 ) P(x) = -\frac{4}{3}(x + 2)(x - 3) P ( x ) = − 3 4 ( x + 2 ) ( x − 3 ) = − 4 3 ( x 2 − x − 6 ) = -\frac{4}{3}(x^2 - x - 6) = − 3 4 ( x 2 − x − 6 ) = − 4 3 x 2 + 4 3 x + 8 = -\frac{4}{3}x^2 + \frac{4}{3}x + 8 = − 3 4 x 2 + 3 4 x + 8
Or in standard form: 4 x 2 − 4 x − 24 = 0 4x^2 - 4x - 24 = 0 4 x 2 − 4 x − 24 = 0 − 3 -3 − 3
Common mistake Mistake 1: Sign Errors with FactorsWrong: Roots are 3 3 3 − 5 -5 − 5 ( x + 3 ) ( x − 5 ) = 0 (x + 3)(x -5) = 0 ( x + 3 ) ( x − 5 ) = 0 Why it feels right: Confusion between "the root is 3" and "the factor involves 3".
The fix: If α \alpha α ( x − α ) (x - \alpha) ( x − α ) ( x + α ) (x + \alpha) ( x + α )
Root = 3 = 3 = 3 = ( x − 3 ) = (x - 3) = ( x − 3 ) x = 3 x = 3 x = 3 = 0 = 0 = 0
Root = − 5 = -5 = − 5 = ( x − ( − 5 ) ) = ( x + 5 ) = (x - (-5)) = (x + 5) = ( x − ( − 5 )) = ( x + 5 ) x = − 5 x = -5 x = − 5 = 0 = 0 = 0
Correct: ( x − 3 ) ( x + 5 ) = 0 (x - 3)(x + 5) = 0 ( x − 3 ) ( x + 5 ) = 0
Common mistake Mistake 2: Forgetting thek k k Wrong: "All quadratics with roots 2 2 2 5 5 5 x 2 − 7 x + 10 = 0 x^2 - 7x + 10 = 0 x 2 − 7 x + 10 = 0 Why it feels right: We learn "the" equation for given roots, implying uniqueness.
The fix: Infinite quadratics share the same roots! They differ by a constant multiple:
x 2 − 7 x + 10 = 0 x^2 - 7x + 10 = 0 x 2 − 7 x + 10 = 0 2 x 2 − 14 x + 20 = 0 2x^2 - 14x + 20 = 0 2 x 2 − 14 x + 20 = 0 − x 2 + 7 x − 10 = 0 -x^2 + 7x - 10 = 0 − x 2 + 7 x − 10 = 0 2 2 2 5 5 5 k ( x 2 − 7 x + 10 ) = 0 k(x^2 - 7x + 10) = 0 k ( x 2 − 7 x + 10 ) = 0 k ≠ 0 k \neq 0 k = 0
Common mistake Mistake 3: Wrong Formula for Sum/ProductWrong: Roots sum to 5 5 5 6 6 6 x 2 + 5 x + 6 = 0 x^2 + 5x + 6 = 0 x 2 + 5 x + 6 = 0
Why it feels right: Directly using the given values.
The fix: The formula is x 2 − ( sum ) x + ( product ) = 0 x^2 -(\text{sum})x + (\text{product}) = 0 x 2 − ( sum ) x + ( product ) = 0 ( x − α ) ( x − β ) = x 2 − ( α + β ) x + α β (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta ( x − α ) ( x − β ) = x 2 − ( α + β ) x + α β Correct: x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 2 2 2 3 3 3 = 5 = 5 = 5 = 6 = 6 = 6
Recall Feynman Explanation (Explain to a 12-year-old)Imagine you're playing a treasure hunt game. You find two treasure chests buried at positions 3 3 3 5 5 5 rule that tells me where the treasures are?"
You could say: "Take any position x x x ( x − 3 ) (x - 3) ( x − 3 ) ( x − 5 ) (x - 5) ( x − 5 ) ( x − 3 ) × ( x − 5 ) (x - 3) \times (x - 5) ( x − 3 ) × ( x − 5 )
Here's the magic: this multiplication equals ZERO exactly when x x x
At x = 3 x = 3 x = 3 ( 3 − 3 ) × ( 3 − 5 ) = 0 × ( − 2 ) = 0 (3 - 3) \times (3 - 5) = 0 \times (-2) = 0 ( 3 − 3 ) × ( 3 − 5 ) = 0 × ( − 2 ) = 0
At x = 5 x = 5 x = 5 ( 5 − 3 ) × ( 5 − 5 ) = 2 × 0 = 0 (5 - 3) \times (5 - 5) = 2 \times 0 = 0 ( 5 − 3 ) × ( 5 − 5 ) = 2 × 0 = 0
Anywhere else: both factors are non-zero, so product is non-zero
When you expand ( x − 3 ) ( x − 5 ) (x - 3)(x - 5) ( x − 3 ) ( x − 5 ) x 2 − 8 x + 15 = 0 x^2 - 8x + 15 = 0 x 2 − 8 x + 15 = 0
Mnemonic Memory Aid: "SUBTRACT the Root"S ubtract the R oot to form the factor.
Root is + 7 +7 + 7 ( x − 7 ) (x \mathbf{-} 7) ( x − 7 ) − 3 -3 − 3 ( x − ( − 3 ) ) = ( x + 3 ) (x \mathbf{-} (-3)) = (x + 3) ( x − ( − 3 )) = ( x + 3 ) For sum/product formula: "S um gets S ubtracted"
x 2 − ( sum ) x + ( product ) = 0 x^2 \mathbf{-} (\text{sum})x + (\text{product}) = 0 x 2 − ( sum ) x + ( product ) = 0
Related concepts:
Vieta's formulas — direct link, this is where sum/product relationships come fromFactor theorem — theoretical basis for why ( x − α ) (x - \alpha) ( x − α ) Quadratic formula — inverse operation: roots → equation vs equation → rootsCompleting the square — alternative way to manipulate quadratic structurePolynomial long division — extends to finding equations from roots for higher-degree polynomialsComplex roots — when α \alpha α β \beta β Graph transformations — the k k k
When you'll use this:
Optimization problems: After finding critical points (roots of derivative), reconstruct the original functionCurve fitting: Given data points where parabola crosses x-axisEngineering: Design projectile paths with specified landing pointsPhysics: Modeling harmonic oscillators with known equilibrium points
#flashcards/maths
What is the general form of a quadratic with roots α and β? :: k ( x − α ) ( x − β ) k(x - \alpha)(x - \beta) k ( x − α ) ( x − β ) k ≠ 0 k \neq 0 k = 0 k [ x 2 − ( α + β ) x + α β ] k[x^2 - (\alpha + \beta)x + \alpha\beta] k [ x 2 − ( α + β ) x + α β ]
If roots of a quadratic are 4 and -3, what is the equation (with k=1)? ( x − 4 ) ( x + 3 ) = x 2 − x − 12 = 0 (x - 4)(x + 3) = x^2 - x - 12 = 0 ( x − 4 ) ( x + 3 ) = x 2 − x − 12 = 0 What is the formula for a quadratic given sum S and product P of roots? x 2 − S x + P = 0 x^2 - Sx + P = 0 x 2 − S x + P = 0 (note the minus sign before S)
Why does the factor form ( x − α ) ( x − β ) (x - \alpha)(x - \beta) ( x − α ) ( x − β ) At
x = α x = \alpha x = α : first factor
( x − α ) = 0 (x - \alpha) = 0 ( x − α ) = 0 , so product is
0 0 0 . At
x = β x = \beta x = β : second factor
( x − β ) = 0 (x - \beta) = 0 ( x − β ) = 0 , so product is
0 0 0 .
If sum of roots is 10 and product is 21, form the equation :: x 2 − 10 x + 21 = 0 x^2 - 10x + 21 = 0 x 2 − 10 x + 21 = 0
What does the constant k in k ( x − α ) ( x − β ) k(x - \alpha)(x - \beta) k ( x − α ) ( x − β ) Vertical scaling of the parabola — determines how "stretched" or "compressed" it is, but doesn't change the root locations
If roots are 3 + 5 3 + \sqrt{5} 3 + 5 3 − 5 3 - \sqrt{5} 3 − 5 Sum =
6 6 6 , Product =
( 3 ) 2 − ( 5 ) 2 = 9 − 5 = 4 (3)^2 - (\sqrt{5})^2 = 9 - 5 = 4 ( 3 ) 2 − ( 5 ) 2 = 9 − 5 = 4 . Equation:
x 2 − 6 x + 4 = 0 x^2 - 6x + 4 = 0 x 2 − 6 x + 4 = 0 Common mistake: If root is -7, what is the factor? ( x − ( − 7 ) ) = ( x + 7 ) (x - (-7)) = (x + 7) ( x − ( − 7 )) = ( x + 7 ) , NOT
( x − 7 ) (x - 7) ( x − 7 ) . Remember: subtract the root.
From Vieta's formulas, if 2 x 2 − 8 x + 6 = 0 2x^2 - 8x + 6 = 0 2 x 2 − 8 x + 6 = 0 c a = 6 2 = 3 \frac{c}{a} = \frac{6}{2} = 3 a c = 2 6 = 3 Why are there infinitely many quadratics with the same two roots? Any non-zero multiple k gives a different quadratic:
k ( x − α ) ( x − β ) k(x - \alpha)(x - \beta) k ( x − α ) ( x − β ) all have the same roots but different shapes/scales
Root alpha: P of alpha = 0
x-alpha and x-beta factors
P = k times x-alpha times x-beta
Form quadratic from sum and product
x^2 - sum x + product = 0
Intuition Hinglish mein samjhoDekho, yeh bahut simple concept hai lekin bohot powerful. Jab tumhe do roots diye ho ek quadratic ke, matlab tumhe pata hai ki parabola X-axis ko kaha kaha touch kar raha hai, toh tum pora equation bana sakte ho backwards se!
Logic yeh hai: agar α ek root hai, matlab x = α pe equation zero ho jati hai. Toh Factor Theorem se, (x - α) ek factor hoga. Dusra root β hai toh (x - β) bhi factor hai. Dono ko multiply karo: P(x) = k(x - α)(x - β). Yaha k ek constant hai jo parabola ki "width" control karta hai — same roots, lekin different shapes. Isko expand karo toh standard form mil jata hai: x² - (sum of roots)x + (product of roots) = 0.
Iska matlab hai ki agar tumhe sirf sum aur product pata ho (Vieta's formulas se), toh bhi equation bana sakte ho bina actual roots jane! Exams mein yeh trick bohot kaam ati hai. Bas ek galti se bacho: root +3 hai toh factor (x - 3) hoga, not (x + 3). Sign confuse mat hona. Practice karo different examples pe, radicals wale bhi, aur samajh mein aa jayega ki kaise roots se equation "reverse engineer" hoti hai.
