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By Shailendra Singh
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Updated on 12 Nov 2025, 15:10 IST
A quadratic equation is a polynomial equation of degree two. When a quadratic polynomial f(x) = ax² + bx + c (where a ≠ 0) is equated to zero, we get a quadratic equation: ax² + bx + c = 0.
Quadratic equations appear frequently in mathematics, physics, engineering, economics, and everyday problem-solving. Understanding how to solve them and interpret their solutions is fundamental for Class 10 CBSE Mathematics.
The general form of a quadratic equation is:
ax² + bx + c = 0
where:
| Type | Condition | Form | Example |
| Pure Quadratic | b = 0, c ≠ 0 | ax² + c = 0 | 3x² - 12 = 0 |
| Incomplete Quadratic | b ≠ 0, c = 0 | ax² + bx = 0 | 5x² + 10x = 0 |
| Simple Quadratic | b = 0, c = 0 | ax² = 0 | 7x² = 0 |
| Complete/Mixed Quadratic | b ≠ 0, c ≠ 0 | ax² + bx + c = 0 | 2x² - 5x + 3 = 0 |
A value x = α is called a root (or zero or solution) of the quadratic equation ax² + bx + c = 0 if it satisfies the equation, meaning aα² + bα + c = 0.
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Solving a quadratic equation means finding all its roots—the values of x that make the equation true.
This is one of the most common methods for solving quadratic equations, especially when the equation can be easily factored.
Solve 5x² + 16x + 12 = 0 by factorization
Solution:
5x² + 16x + 12 = 0 5x² + 10x + 6x + 12 = 0 [Split middle term: 16x = 10x + 6x] 5x(x + 2) + 6(x + 2) = 0 [Factor by grouping] (x + 2)(5x + 6) = 0 [Common factor (x + 2)]
Therefore: x = -2 or x = -6/5
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This method transforms the quadratic equation into a perfect square form.
Solve x² - (√3 + 1)x + √3 = 0 by completing the square
Solution:
x² - (√3 + 1)x = -√3 x² - 2[(√3 + 1)/2]x + [(√3 + 1)/2]² = -√3 + [(√3 + 1)/2]² [x - (√3 + 1)/2]² = (√3 - 1)²/4 x - (√3 + 1)/2 = ±(√3 - 1)/2 x = (√3 + 1)/2 ± (√3 - 1)/2
Therefore: x = √3 or x = 1
This is the most powerful and universal method for solving any quadratic equation.
Starting with: ax² + bx + c = 0 (a ≠ 0)
This is the quadratic formula.
While it's essential to understand the manual methods, quadratic equation calculators can:
Recommended approach: Always solve manually first, then verify with a calculator.
| Formula Name | Mathematical Expression | Explanation |
| Quadratic Formula | x = [-b ± √(b² - 4ac)]/2a | Universal formula for solving ax² + bx + c = 0 |
| Discriminant | D = b² - 4ac | Determines nature of roots |
| Sum of Roots | α + β = -b/a | Relationship between roots and coefficients |
| Product of Roots | αβ = c/a | Relationship between roots and coefficients |
| Equation from Roots | x² - (α + β)x + αβ = 0 | Forms equation when roots are known |
| Roots Formula | α = [-b + √D]/2a, β = [-b - √D]/2a | Individual root expressions |
The expression D = b² - 4ac is called the discriminant of the quadratic equation ax² + bx + c = 0.
The discriminant tells us everything about the nature of roots without actually solving the equation:
| Discriminant Value | Nature of Roots | Details |
| D > 0 | Real and Distinct | Two different real roots |
| D = 0 | Real and Equal | Two identical real roots (α = β = -b/2a) |
| D < 0 | Imaginary/Complex | No real roots (roots are complex conjugates) |
If D > 0:
Find the nature of roots of 2x² + 7x + 5 = 0
Solution:
D = b² - 4ac D = 7² - 4(2)(5) D = 49 - 40 = 9 Since D > 0 and D = 9 (perfect square) The roots are real, rational, and unequal.
For the quadratic equation ax² + bx + c = 0 with roots α and β:
If α and β are roots, the quadratic equation is: x² - (α + β)x + αβ = 0
Or equivalently: x² - (sum of roots)x + (product of roots) = 0
For ax² + bx + c = 0:
Factor 6x² + 13x + 5 = 0
Solution:
ac = 6 × 5 = 30 Factors of 30: (1,30), (2,15), (3,10), (5,6) Which pair sums to 13? → 3 + 10 = 13 ✓ 6x² + 3x + 10x + 5 = 0 3x(2x + 1) + 5(2x + 1) = 0 (2x + 1)(3x + 5) = 0 Roots: x = -1/2 or x = -5/3
While the quadratic formula works universally, other methods can be more efficient:
| Method | Best For |
| Factorization | Integer or simple rational roots |
| Quadratic Formula | Non-factorable equations, decimal/complex roots |
| Completing Square | Finding vertex, deriving formulas |
| Graphical | Visualization, approximate solutions |
Many non-quadratic equations can be solved by substitution to convert them into quadratic form.
Form: ax⁴ + bx² + c = 0
Method: Substitute x² = y to get ay² + by + c = 0
Example: Solve 2x⁴ - 5x² + 3 = 0
Let x² = y 2y² - 5y + 3 = 0 (2y - 3)(y - 1) = 0 y = 3/2 or y = 1 Therefore: x² = 3/2 → x = ±√(3/2) x² = 1 → x = ±1
Form: a[x² + 1/x²] + b[x + 1/x] + c = 0
Method:
Example: Solve 9[x² + 1/x²] - 9[x + 1/x] - 52 = 0
Let x + 1/x = y Then x² + 1/x² = y² - 2 9(y² - 2) - 9y - 52 = 0 9y² - 9y - 70 = 0 y = 10/3 or y = -7/3
Form: (x+a)(x+b)(x+c)(x+d) + k = 0, where a+b = c+d
Method: Regroup as [(x+a)(x+b)][(x+c)(x+d)] + k = 0, then substitute
Example: Solve (x+1)(x+2)(x+3)(x+4) = 120
Note: 1+4 = 2+3 = 5 [(x+1)(x+4)][(x+2)(x+3)] = 120 (x² + 5x + 4)(x² + 5x + 6) = 120 Let x² + 5x = y (y+4)(y+6) = 120 y² + 10y - 96 = 0 y = 6 or y = -16 x² + 5x = 6 → x = 1 or x = -6 x² + 5x = -16 → No real solution
Quadratic equations are essential for solving practical problems in various fields.
Problem: The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the dimensions.
Solution:
Let shorter side = x meters Longer side = (x + 30) meters Diagonal = (x + 60) meters By Pythagorean theorem: x² + (x+30)² = (x+60)² x² + x² + 900 + 60x = x² + 3600 + 120x x² - 60x - 2700 = 0 (x - 90)(x + 30) = 0 x = 90 meters (rejecting negative value) Dimensions: 90 m × 120 m
Problem: A train travels 360 km at uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed.
Solution:
Let speed = x km/h Time taken = 360/x hours With increased speed: Time = 360/(x+5) hours Given: 360/x - 360/(x+5) = 1 360(x+5) - 360x = x(x+5) 1800 = x² + 5x x² + 5x - 1800 = 0 (x + 45)(x - 40) = 0 x = 40 km/h (rejecting negative value)
Problem: Seven years ago, Varun's age was five times the square of Swati's age. Three years hence, Swati's age will be two-fifths of Varun's age. Find their present ages.
Solution:
Let present ages: Varun = x years, Swati = y years Seven years ago: x - 7 = 5(y - 7)² x = 5y² - 70y + 252 ... (1) Three years hence: y + 3 = (2/5)(x + 3) 5y + 15 = 2x + 6 x = (5y + 9)/2 ... (2) From (1) and (2): 5y² - 70y + 252 = (5y + 9)/2 10y² - 145y + 495 = 0 2y² - 29y + 99 = 0 (y - 9)(2y - 11) = 0 y = 9 years (valid solution) x = 27 years Present ages: Varun = 27 years, Swati = 9 years
The graph of y = ax² + bx + c is always a parabola.
| When a > 0 | When a < 0 |
| Parabola opens upward (∪) | Parabola opens downward (∩) |
| Has minimum value | Has maximum value |
| Vertex is the lowest point | Vertex is the highest point |
For a > 0:
For a < 0: Signs reverse
| Condition | Result |
| b = 0 | Roots are equal in magnitude but opposite in sign (α = -β) |
| c = 0 | One root is zero, other is -b/a |
| a = c | Roots are reciprocals of each other |
| a > 0, c < 0 or a < 0, c > 0 | Roots have opposite signs |
| a + b + c = 0 | One root is 1, other is c/a |
| a, b, c ∈ Z and roots rational | Roots must be integers |
Mastering quadratic equations is crucial not just for Class 10 board exams, but also for higher mathematics, competitive exams, and real-world problem solving. The key is understanding when to apply which method, recognizing patterns in problems, and practicing systematically.
Remember: Every quadratic equation tells a story through its roots, discriminant, and graph. Learn to read that story, and you'll find these equations become intuitive friends rather than intimidating foes.
Practice regularly, solve diverse problems, and always verify your solutions. Success in quadratic equations comes from understanding, not memorization.
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Use factorization when factors are obvious or coefficients are small. Use the formula for complex coefficients or when factorization isn't apparent.
No. A quadratic equation can have at most 2 roots (fundamental theorem of algebra).
The equation has no real roots, only complex conjugate roots (beyond Class 10 syllabus).
Substitute the roots back into the original equation. If both sides equal, your answer is correct.
Yes! If a + b + c = 0, one root is 1 and the other is c/a. If coefficients are symmetric (a = c), roots are reciprocals.
A quadratic equation is a polynomial equation of degree two in one variable. It takes the standard form ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. The term "quadratic" comes from "quad" meaning square, as the variable is squared. Examples include 2x² - 3x + 1 = 0 and x² - 5x + 6 = 0. These equations appear frequently in real-world problems involving area, projectile motion, profit optimization, and many other applications.
The roots of a quadratic equation are the values of x that satisfy the equation, making it equal to zero. In other words, if you substitute a root into the equation ax² + bx + c = 0, the left side equals zero. A quadratic equation can have two real roots, one repeated real root, or two complex roots depending on the discriminant value. Finding these roots is often called "solving" the quadratic equation.
The discriminant is the expression b² - 4ac from the quadratic formula, typically denoted by D or Δ. It determines the nature of the roots without actually solving the equation. If D > 0, the equation has two distinct real roots; if D = 0, it has two equal real roots (one repeated root); and if D < 0, it has no real roots (two complex conjugate roots). The discriminant is crucial for understanding whether real solutions exist before attempting to solve the equation.
The quadratic formula, also known as Sreedharacharya's formula, provides the solutions to any quadratic equation ax² + bx + c = 0 where a ≠ 0. The formula is: x = (-b ± √(b² - 4ac)) / 2a. This gives two roots: α = (-b + √D) / 2a and β = (-b - √D) / 2a, where D is the discriminant. This universal method works for all quadratic equations and is particularly useful when factorization is difficult or impossible.
Factorization involves expressing the quadratic equation as a product of two linear factors. First, find two numbers whose product equals the product of the coefficient of x² and the constant term, and whose sum equals the coefficient of x. Then split the middle term and group the terms to factor by grouping. For example, x² + 5x + 6 = 0 factors as (x + 2)(x + 3) = 0, giving roots x = -2 and x = -3. This method is efficient when the roots are rational numbers.
Completing the square is a technique that transforms a quadratic equation into a perfect square trinomial. You make the coefficient of x² equal to 1, move the constant to the right side, add the square of half the coefficient of x to both sides, and then take the square root. This method not only solves equations but also helps derive the quadratic formula and is useful in graphing parabolas and optimization problems.
For a quadratic equation ax² + bx + c = 0 with roots α and β, the sum of roots α + β = -b/a (negative of coefficient of x divided by coefficient of x²), and the product of roots αβ = c/a (constant term divided by coefficient of x²). These relationships allow you to find information about roots without solving the equation completely, form new equations with given roots, and verify your solutions.
If you know the roots α and β of a quadratic equation, you can form the equation using: x² - (α + β)x + αβ = 0, or more generally, a[x² - (sum of roots)x + product of roots] = 0. For example, if the roots are 2 and 3, the equation is x² - 5x + 6 = 0. This concept is particularly useful in problems where you need to construct equations based on conditions about the roots.
Quadratic equations model numerous practical situations including calculating areas of rectangular plots, determining projectile motion trajectories, optimizing profit in business, analyzing age-related problems, computing travel time and speed relationships, designing structures with specific dimensions, and solving problems involving consecutive numbers. Any scenario involving squared terms or relationships between quantities that follow a parabolic pattern can potentially be modeled using quadratic equations.
A quadratic equation has equal (or coincident) roots when the discriminant D = b² - 4ac equals zero. In this case, both roots have the same value: α = β = -b/2a. Geometrically, this means the parabola just touches the x-axis at one point (the vertex). For example, x² - 6x + 9 = 0 has equal roots because D = 36 - 36 = 0, and both roots equal 3.
A quadratic equation ax² + bx + c = 0 has real roots if and only if the discriminant D = b² - 4ac ≥ 0. If D > 0, there are two distinct real roots; if D = 0, there are two equal real roots. If D < 0, the equation has no real roots but has two complex conjugate roots. Checking the discriminant before attempting to solve can save time and help you understand the nature of solutions.
When a quadratic equation with rational coefficients has a positive discriminant that is a perfect square, the roots are rational numbers. If the discriminant is positive but not a perfect square, the roots are irrational and come in conjugate pairs like p + √q and p - √q, where p is rational and √q is a surd. For example, x² - 2x - 1 = 0 has irrational roots (1 + √2) and (1 - √2).
Many equations that don't initially appear quadratic can be solved by substitution. For example, x⁴ - 5x² + 4 = 0 becomes quadratic by letting y = x². Similarly, equations like (x² + 3x)² - (x² + 3x) - 6 = 0 can be simplified by substituting y = x² + 3x. After solving for y, substitute back to find x. This technique extends the power of quadratic methods to more complex equations.
For a quadratic equation ax² + bx + c = 0 to have both roots negative, three conditions must be satisfied: the discriminant must be non-negative (D ≥ 0), the sum of roots must be negative (α + β < 0), and the product of roots must be positive (αβ > 0). This translates to: b² - 4ac ≥ 0, -b/a < 0, and c/a > 0, which means a, b, and c must all have the same sign.
In word problems, first translate the situation into mathematical expressions by defining variables. Set up equations based on the given conditions, which often leads to a quadratic equation. Solve using the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. Finally, interpret the solutions in the context of the problem, rejecting any that don't make physical sense (like negative lengths or times). Always verify your answer against the original problem conditions.