Animated Diagram — Quadratic Equations

a = 1

b = 0

c = 0

Standard form: ax² + bx + c = 0 | Vertex: (0, 0) | Roots: x = 0 (repeated)

💡 Drag sliders to see how a (width/direction), b (tilt), and c (vertical shift) change the parabola. Roots are where the curve crosses the x-axis.

Solving by Factorisation (Splitting the Middle Term)

Example: x² − 5x + 6 = 0

Step 1

Identify a = 1, b = −5, c = 6. Product ac = 1 × 6 = 6.

Step 2

Find two numbers whose product = ac = 6 and sum = b = −5.
Numbers: −2 and −3 (since −2 × −3 = 6, −2 + (−3) = −5).

Step 3

Split the middle term: x² − 2x − 3x + 6 = 0

Step 4

Group: (x² − 2x) + (−3x + 6) = 0

Step 5

Factor each group: x(x − 2) − 3(x − 2) = 0

Step 6

Common factor: (x − 2)(x − 3) = 0

Step 7

Apply zero-product rule: x − 2 = 0 or x − 3 = 0

Step 8

∴ Roots: x = 2 or x = 3
Verify: 2² − 5(2) + 6 = 0 ✓ 3² − 5(3) + 6 = 0 ✓

Solving by Completing the Square

Example: 2x² − 8x + 3 = 0

Step 1

Divide by a = 2: x² − 4x + 3/2 = 0

Step 2

Move constant to RHS: x² − 4x = −3/2

Step 3

Half of coefficient of x: (−4)/2 = −2. Square it: (−2)² = 4

Step 4

Add 4 to both sides: x² − 4x + 4 = −3/2 + 4 = 5/2

Step 5

Write LHS as perfect square: (x − 2)² = 5/2

Step 6

Take square root: x − 2 = ±√(5/2)

Step 7

∴ x = 2 ± √(5/2) = 2 ± √10/2
x ≈ 3.58 or x ≈ 0.42

💡 The Quadratic Formula is actually derived from this method — applied to the general equation ax² + bx + c = 0.

x = (−b ± √(b² − 4ac)) / 2a
The Shreedharacharya Formula — works for ALL quadratic equations

Quadratic Equation Solver

Click Solve to see step-by-step solution using the quadratic formula.

Discriminant D = b² − 4ac determines the nature of roots

✌️

Two Real & Distinct

D > 0 (b² − 4ac > 0)

Parabola crosses x-axis at two points. Roots are x = (−b ± √D)/2a

🤝

Two Equal (Repeated)

D = 0 (b² − 4ac = 0)

Parabola touches x-axis at exactly one point. Both roots = −b/2a

🚫

No Real Roots

D < 0 (b² − 4ac < 0)

Parabola doesn't cross x-axis. Roots are imaginary (not in real numbers).

Example: x² − 5x + 6 = 0 → D = 25 − 24 = 1 > 0 → Two distinct real roots (x = 2, 3)

💡 Click any problem type to see how it converts to a quadratic equation.

🔢Number Problems

Pattern: Two numbers with given sum/difference and product.
Let numbers = x, (S − x). Then x(S − x) = P → x² − Sx + P = 0 Example: Two numbers sum to 15, product is 56. Find them.
x(15 − x) = 56 → x² − 15x + 56 = 0 → (x − 7)(x − 8) = 0
Numbers: 7 and 8

🔲Area & Geometry

Pattern: Dimensions expressed in terms of one variable, area given.
Rectangle: length × breadth = Area → x(x + d) = A Example: Length is 5 more than breadth. Area = 150 m².
x(x + 5) = 150 → x² + 5x − 150 = 0 → (x + 15)(x − 10) = 0
Breadth = 10 m, Length = 15 m (reject negative)

⏱️Speed, Distance & Time

Pattern: D = S × T with one variable, usually gives (x)(x ± k) form.
If speed increases by k: D/x − D/(x+k) = T_diff Example: Train travels 480 km. If speed ↑ by 8 km/h, takes 3 hrs less.
480/x − 480/(x+8) = 3 → x² + 8x − 1280 = 0
Original speed = 32 km/h

🧑‍🤝‍🧑Age Problems

Pattern: Product of ages (present/past/future) is given.
Present × Past_or_Future = Product → x(x ± k) = P Example: Product of Ravi's age 5 years ago and 3 years later is 51.
(x − 5)(x + 3) = 51 → x² − 2x − 66 = 0
x = (2 + √268)/2 ≈ 9.2 → x = 9 (approx)

👷Work & Time

Pattern: Combined rates with quadratic relationship.
If A does in x days and B in (x+k) days: 1/x + 1/(x+k) = 1/T Example: Two pipes fill a tank: Pipe A takes 5 hrs less than B. Together they fill in 6 hrs.
1/x + 1/(x+5) = 1/6 → x² − 7x − 30 = 0 → (x−10)(x+3) = 0
Pipe B = 10 hrs, Pipe A = 15 hrs... wait let me recheck.
Actually: Pipe A = 10 hrs, Pipe B = 15 hrs (reject negative)

🔗Consecutive Numbers

Pattern: Sum of squares, product, or other relations of consecutive integers.
Consecutive: x, x+1 | Even: x, x+2 | Odd: x, x+2 Example: Sum of squares of two consecutive odd numbers is 290.
x² + (x+2)² = 290 → 2x² + 4x + 4 = 290 → x² + 2x − 143 = 0
(x + 13)(x − 11) = 0 → Numbers: 11 and 13

📐Pythagoras & Triangles

Pattern: Sides expressed algebraically, use h² = p² + b².
If sides are x, x+a, x+b: (x+b)² = x² + (x+a)² Example: Hypotenuse is 2 more than one side, which is 1 more than the other.
Sides: x, x+1, x+2: (x+2)² = x² + (x+1)²
x² + 4x + 4 = x² + x² + 2x + 1 → x² − 2x − 3 = 0
(x−3)(x+1) = 0 → Sides: 3, 4, 5

💰Profit & Cost

Pattern: Revenue = price × quantity; with price/demand relationship.
If price = (base + x), items sold = (base − x): Revenue = (a+x)(b−x) Example: Cost of each item is ₹(x + 5). Number sold = (50 − x). Total revenue = ₹750.
(x+5)(50−x) = 750 → −x² + 45x + 250 = 750
x² − 45x + 500 = 0 → (x−20)(x−25) = 0
x = 20 or x = 25