CBSE Class 10 Maths Chapter Quadratic Equations Notes

Quadratic Equations is one of the most important chapters in CBSE Class 10 Maths because it introduces students to equations of degree two and their practical applications. In this chapter, students learn how to identify a quadratic equation, write it in the standard form, and solve it using different methods such as factorisation, completing the square, and the quadratic formula. The chapter builds a strong foundation for higher mathematics and helps students develop logical and analytical thinking skills. Understanding quadratic equations is essential for solving problems related to geometry, physics, business mathematics, and real-life situations involving area, speed, and dimensions. 

Students also learn about the nature of roots and the relationship between the roots and coefficients of a quadratic equation. The concepts covered in this chapter frequently appear in school examinations, board exams, scholarship tests, and competitive entrance assessments. Well-prepared notes for CBSE Class 10 Maths Chapter Quadratic Equations help students revise formulas, important concepts, solved examples, and examination-oriented questions quickly. By mastering this chapter, students gain confidence in algebraic problem-solving and improve their overall mathematical accuracy. Regular practice of quadratic equation questions ensures better conceptual clarity and helps students score high marks in the CBSE Class 10 Mathematics examination. Also check out NCERT Exemplar for Class 10 Maths, NCERT Solutions for Class 10, & Maths formulas prepared and developed by experts at Myclass24. 

Understanding Quadratic Equations: Definitions, Solution Methods, and Applications

What is a Quadratic Equation?

A quadratic equation is a polynomial equation of degree 2, expressed in the general form ax² + bx + c = 0, where a ≠ 0 and a, b, c are real numbers. The coefficient 'a' must be non-zero; otherwise, the equation reduces to a linear form. Quadratic equations are fundamental in algebra and appear extensively in physics, engineering, economics, and various scientific applications.

Quadratic equations are classified into two main categories:

1. Pure Quadratic Equations: These take the form ax² + c = 0 (where b = 0), containing only the squared term and constant.
2. Affected Quadratic Equations: These follow the complete form ax² + bx + c = 0 (where b ≠ 0), including all three terms.

The solutions to a quadratic equation are called roots or zeros of the polynomial. A quadratic equation can have at most two roots, which may be real or complex depending on the discriminant value.

The Discriminant: Determining the Nature of Roots

The discriminant (denoted as D or Δ) is a critical component in analyzing quadratic equations, calculated as D = b² - 4ac. This value determines whether roots are real, equal, or imaginary:

• D > 0: The equation has two distinct real roots. If D is a perfect square and coefficients are rational, roots are rational; otherwise, they are irrational and occur in conjugate pairs.
• D = 0: The equation has two equal real roots (also called a repeated root), calculated as -b/2a.
• D < 0: The equation has no real roots; instead, it has two complex conjugate roots of the form a + bi and a - bi.

Understanding the discriminant is essential for predicting solution characteristics before actual computation, making it a powerful analytical tool in higher mathematics.

Methods for Solving Quadratic Equations

1. Factorization Method

This method involves expressing the quadratic equation as a product of two linear factors. The process includes:

• Identifying factors of the constant term (c) whose sum or difference equals the coefficient of x (b)
• Splitting the middle term based on these factors
• Grouping terms and factoring out common elements
• Setting each factor equal to zero to find the roots

Factorization is most effective when roots are rational numbers and the equation can be easily decomposed.

2. Completing the Square Method

This technique transforms the quadratic equation into a perfect square trinomial:

• Ensure the coefficient of x² is 1 (divide through by 'a' if necessary)
• Move the constant term to the right side
• Add the square of half the coefficient of x to both sides
• Express the left side as a perfect square
• Take the square root of both sides and solve for x

This method provides geometric insight and is particularly useful for deriving the quadratic formula.

3. Quadratic Formula (Sridharacharya's Formula)

The most universal method applicable to all quadratic equations is the quadratic formula:

x = [-b ± √(b² - 4ac)] / 2a

This formula, attributed to the ancient Indian mathematician Sridharacharya, directly yields both roots regardless of whether they are rational, irrational, or complex. It's derived through completing the square on the general form.

Consolidated Formula Reference Table

Real-World Applications of Quadratic Equations

Quadratic equations model numerous practical scenarios:

1. Motion and Projectile Problems: Calculating the trajectory of objects under gravity follows quadratic relationships, such as determining maximum height or range.

2. Area and Perimeter Optimization: Finding dimensions of rectangles, triangles, or other shapes when given area constraints involves quadratic formulations.

3. Business and Economics: Profit maximization, cost minimization, and break-even analysis often require solving quadratic equations relating revenue, costs, and quantities.

4. Number Problems: Many classical mathematical puzzles involving consecutive integers, digit relationships, or reciprocals lead to quadratic equations.

5. Speed, Distance, and Time: Problems involving relative motion, such as boats in streams or vehicles at varying speeds, generate quadratic relationships.

Important Properties and Relationships

When working with quadratic equations, several key relationships prove valuable:

• If a quadratic equation has more than two distinct solutions, it must be an identity (true for all x values)
• For equations with rational coefficients, irrational roots always occur in conjugate pairs (like 2 + √3 and 2 - √3)
• Complex roots with real coefficients always appear as conjugate pairs (such as 3 + 4i and 3 - 4i)
• The axis of symmetry of the parabola represented by the quadratic equation is x = -b/2a
• Sign analysis: If a and c have the same sign but opposite to b, both roots are positive; if a, b, and c all have the same sign, both roots are negative

Solving Strategy and Common Pitfalls

When approaching quadratic equations, follow this systematic approach:

1. Ensure the equation is in standard form (ax² + bx + c = 0)
2. Calculate the discriminant to predict root nature
3. Choose the most efficient solution method based on the equation structure
4. Verify solutions by substituting back into the original equation
5. Reject extraneous solutions that don't satisfy domain restrictions

Common mistakes include forgetting that a ≠ 0, errors in sign when applying the quadratic formula, and accepting negative solutions in contexts where only positive values are meaningful (such as length, time, or quantity problems).

A polynomial of degree 2 (i.e., ax² + bx + c) is called a quadratic polynomial where a ≠ 0 and a, b, c are real numbers.

General Form of a Quadratic Equation

ax² + bx + c = 0
where a, b, c are real numbers and a ≠ 0
Since a ≠ 0, quadratic equations, in general, are of the following types:

1. b = 0, c ≠ 0 i.e., of the type ax² + c = 0
2. b ≠ 0, c = 0 i.e., of the type ax² + bx = 0
3. b = 0, c = 0 i.e., of the type ax² = 0
4. b ≠ 0, c ≠ 0 i.e., of the type ax² + bx + c = 0

Classification of Quadratic Equations

Quadratic equations are classified into two categories:

1. Pure Quadratic Equation

Of the form ax² + c = 0 (i.e., b = 0 in ax² + bx + c = 0)

Examples:x² - 4 = 0 and 3x² + 1 = 0

2. Affected Quadratic Equation

Of the form ax² + bx + c = 0, b ≠ 0

Examples:x² + 2x - 8 = 0 and 5x² + 3x - 2 = 0

Zeros of Quadratic Polynomial

For a quadratic polynomial p(x) = ax² + bx + c, those values of x for which ax² + bx + c = 0 is satisfied, are called zeros of quadratic polynomial p(x).

i.e., if p(α) = aα² + bα + c = 0, then α is called the zero of quadratic polynomial.

Roots of a Quadratic Equation

The value of x which satisfies the given quadratic equation is known as its root. The roots of the given equation are known as its solution.

Derivation of Quadratic Formula

General form of a quadratic equation is: ax² + bx + c = 0

Multiplying by 4a: 4a²x² + 4abx + 4ac = 0

Rearranging: 4a²x² + 4abx = -4ac

Adding b² both sides: 4a²x² + 4abx + b² = b² - 4ac

Factoring: (2ax + b)² = b² - 4ac

Taking square root: 2ax + b = ±√(b² - 4ac)

x = (-b ± √(b² - 4ac)) / 2a

Remark:

A quadratic equation is satisfied by exactly two values of 'x' which may be real or imaginary.

• A quadratic equation if a ≠ 0 → Two roots
• A linear equation if a = 0, b ≠ 0 → One root
• A contradiction if a = b = 0, c ≠ 0 → No root
• An identity if a = b = c = 0 → Infinite roots

Important: A quadratic equation cannot have more than two roots.

Nature of Roots

Consider the quadratic equation ax² + bx + c = 0 having α and β as its roots and b² - 4ac is called discriminant of roots of quadratic equation. It is denoted by D or Δ.

Cases Based on Discriminant

Case 1: When b² - 4ac > 0 (D > 0)

In this case roots of the given equation are real and distinct:
α = (-b + √(b² - 4ac)) / 2a and β = (-b - √(b² - 4ac)) / 2a
Sub-cases:

• When a(≠ 0), b, c ∈ Q and b² - 4ac is a perfect square: Both roots are rational and distinct
• When a(≠ 0), b, c ∈ Q and b² - 4ac is not a perfect square: Both roots are irrational and distinct

Case 2: When b² - 4ac = 0 (D = 0)

In this case both the roots are real and equal to -b/2a

Case 3: When b² - 4ac < 0 (D < 0)

In this case b² - 4ac < 0, then 4ac - b² > 0

The roots are:
α = (-b + i√(4ac - b²)) / 2a and β = (-b - i√(4ac - b²)) / 2a
i.e., in this case both roots are imaginary and distinct.

Remark:

1. If a,b,c Q and b² - 4ac is positive (D > 0) but not a perfect square, then the roots are irrational and they always occur in conjugate pairs like and . However, if a, b, c are irrational number and b² – 4ac is positive but not a perfect square, then the roots may not occur in conjugate pairs.
2. If  b² – 4ac is negative (D > 0), then the roots are complex conjugate of each other. In fact, complex roots of an equation with real coefficients always occur in conjugate pairs like 2 + 3i and 2 – 3i. However, this may not be true in case of equations with complex coefficients. For example, x² – 2ix – 1 = 0 has both roots equal to i.
3. If a and c are of the same sign and b has a sign opposite to that of a as well as c, then both the roots are positive, the sum as well as the product of roots is positive .
4. If a,b, are of the same sign then both the roots are negative, the sum of the roots is negative but the product of roots is positive .

Methods of Solving Quadratic Equations

Method 1: By Factorization

Algorithm:

1. Factorize the constant term of the given quadratic equation
2. Express the coefficient of middle term as the sum or difference of the factors obtained in step 1
3. Split the middle term in two parts obtained in step 2
4. Factorize the quadratic equation obtained in step 3

Example 1: Solve x² - 2ax + a² - b² = 0

Here, Factors of constant term (a² - b²) are (a - b) and (a + b)

Also, Coefficient of the middle term = -2a = -[(a - b) + (a + b)]

∴ x² - 2ax + a² - b² = 0

⟹ x² - {(a - b) + (a + b)}x + (a - b)(a + b) = 0

⟹ x² - (a - b)x - (a + b)x + (a - b)(a + b) = 0

⟹ x{x - (a - b)} - (a + b){x - (a - b)} = 0

⟹ {x - (a - b)}{x - (a + b)} = 0

⟹ x - (a - b) = 0 or x - (a + b) = 0

⟹ x = a - b or x = a + b

Example 2: Solve 64x² - 625 = 0

We have 64x² - 625 = 0

or (8x)² - (25)² = 0

or (8x + 25)(8x - 25) = 0

i.e., 8x + 25 = 0 or 8x - 25 = 0

This gives x = -25/8 or x = 25/8

Thus, x = ±25/8 are solutions of the given equation.

Method 2: By Completion of Square

Algorithm:

1. Obtain the quadratic equation. Let it be ax² + bx + c = 0, a ≠ 0
2. Make the coefficient of x² unity, if it is not unity. i.e., obtain x² + (b/a)x + c/a = 0
3. Shift the constant term c/a on R.H.S. to get x² + (b/a)x = -c/a
4. Add square of half of the coefficient of x i.e., (b/2a)² on both sides
5. Write L.H.S. as the perfect square of a binomial expression and simplify R.H.S.
6. Take square root of both sides
7. Obtain the values of x by shifting the constant term on RHS

Example: Solve x² + 3x + 1 = 0

We have x² + 3x + 1 = 0

Add and subtract (1/2 × coefficient of x)² in L.H.S.:

x² + 3x + (3/2)² - (3/2)² + 1 = 0

(x + 3/2)² - 9/4 + 1 = 0

(x + 3/2)² = 9/4 - 1 = 5/4

x + 3/2 = ±√(5/4) = ±√5/2

This gives x = -3/2 ± √5/2

Therefore x = (-3 ± √5)/2 are the solutions of the given equation.

Method 3: Using Quadratic Formula

Steps:

1. By comparison with general quadratic equation, find the value of a, b and c
2. Find the discriminant of the quadratic equation: D = b² - 4ac
3. Find the roots using: x = (-b ± √D)/2a

Remark:

If b² - 4ac < 0 i.e., negative, then √(b² - 4ac) is not real and therefore, the equation does not have any real roots.

Example: Solve the quadratic equation x² - 7x - 5 = 0

Comparing the given equation with ax² + bx + c = 0, we find that a = 1, b = -7 and c = -5.

Therefore, D = (-7)² - 4 × 1 × (-5) = 49 + 20 = 69 > 0

Since D is positive, the equation has two roots given by:

x = (7 ± √69)/2

Therefore x = (7 + √69)/2, (7 - √69)/2 are the required solutions.

Example: For what value of k, (4 – k)x² + (2k + 4)x + (8k + 1) is a perfect square.

Sol. The given equation is a perfect square, if its discriminate is zero i.e. (2k + 4)² – 4(4 - k) (8k + 1) = 0

4(k + 2)² – 4(4 – k) (8k + 1) = 0 Þ 4[4(k + 2)² – (4 – k) (8k + 1)] = 0

[(k² + 4k + 4) – (–8k² + 31k + 4)] = 0 Þ 9k² – 27k = 0

9k (k – 3) = 0 Þ k = 0 or k = 3

Hence, the given equation is a perfect square, if k = 0 or k = 3.

Example 1: The sum of the squares of two consecutive positive integers is 545. Find the integers.

Let x be one of the positive integers. Then the other integer is x + 1, x ∈ Z⁺

Since the sum of the squares of the integers is 545, we get:

x² + (x + 1)² = 545

or 2x² + 2x - 544 = 0

or x² + x - 272 = 0

x² + 17x - 16x - 272 = 0

or x(x + 17) - 16(x + 17) = 0

or (x - 16)(x + 17) = 0

Here, x = 16 or x = -17. But, x is a positive integer. Therefore, reject x = -17 and take x = 16.

Hence, two consecutive positive integers are 16 and (16 + 1), i.e., 16 and 17.

Example 2: The length of a hall is 5 m more than its breadth. If the area of the floor of the hall is 84 m², what are the length and the breadth of the hall?

Let the breadth of the hall be x metres.

Then the length of the hall is (x + 5) metres.

The area of the floor = x(x + 5) m²

Therefore, x(x + 5) = 84

or x² + 5x - 84 = 0

or (x + 12)(x - 7) = 0

This gives x = 7 or x = -12.

Since, the breadth of the hall cannot be negative, we reject x = -12 and take x = 7 only.

Thus, breadth of the hall = 7 metres, and length of the hall = (7 + 5), i.e., 12 metres.

Problem 1: In each of the following, determine whether the given values of x are the solutions of the given equation or not:

(i) x² - 6x + 5 = 0; x = -1, x = -5

(ii) 6x² - x - 2 = 0; x = -1/2, x = 2/3

Solution:

(i) Consider p(x) = x² - 6x + 5 ...(i)

Substituting x = -1 in (i), we get:

p(-1) = (-1)² - 6(-1) + 5 = 1 + 6 + 5 = 12 ≠ 0

Hence x = -1 is not the solution of given equation.

Again substituting x = -5 in (i), we get:

p(-5) = (-5)² - 6(-5) + 5 = 25 + 30 + 5 = 60 ≠ 0

Hence x = -5 is not the solution of given equation.

Therefore, x = -1 and x = -5 are NOT solutions of the given equation.

(ii) Consider p(x) = 6x² - x - 2 ...(i)

Substituting x = -1/2 in (i), we get:

p(-1/2) = 6(-1/2)² - (-1/2) - 2 = 6(1/4) + 1/2 - 2 = 3/2 + 1/2 - 2 = 2 - 2 = 0

Hence x = -1/2 is the solution of given equation.

Again substituting x = 2/3 in (i), we get:

p(2/3) = 6(2/3)² - (2/3) - 2 = 6(4/9) - 2/3 - 2 = 8/3 - 2/3 - 2 = 6/3 - 2 = 2 - 2 = 0

Hence x = 2/3 is the solution of given equation.

Therefore, x = -1/2 and x = 2/3 are solutions of the quadratic equation.

Problem 2: Determine whether the given values of x are solutions of the given equation or not:

x² + 2√3x - 9 = 0; x = √3, x = -3√3

Solution: Consider p(x) = x² + 2√3x - 9 ...(i)

Substituting x = √3 in (i):

p(√3) = (√3)² + 2√3(√3) - 9 = 3 + 6 - 9 = 0

x = √3 is a solution of the given quadratic equation.

Substituting x = -3√3 in (i) we get:

p(-3√3) = (-3√3)² + 2√3(-3√3) - 9 = 27 - 18 - 9 = 0 = RHS

x = -3√3 is a solution of the given quadratic equation.

Problem 3: If x = 2 and x = 3 are roots of 3x² - 2px + 2q = 0, find the values of p and q.

Solution:

Since x = 2 is a root of the given equation, we have:

3(2)² - 2p(2) + 2q = 0

12 - 4p + 2q = 0

4p - 2q = 12 ...(i)

Since x = 3 is a root of the given equation, we have:

3(3)² - 2p(3) + 2q = 0

27 - 6p + 2q = 0

6p - 2q = 27 ...(ii)

Subtracting equation (i) from equation (ii), we get:

6p - 2q - 4p + 2q = 27 - 12

2p = 15

p = 15/2

Substituting p = 15/2 in equation (i), we get:

4(15/2) - 2q = 12

30 - 2q = 12

2q = 18

q = 9

Hence p = 15/2 and q = 9.

Problem 4: Solve 64x² - 625 = 0

Solution:

We have 64x² - 625 = 0

or (8x)² - (25)² = 0

or (8x + 25)(8x - 25) = 0

i.e., 8x + 25 = 0 or 8x - 25 = 0

This gives x = -25/8 or x = 25/8

Thus, x = ±25/8 are solutions of the given equation.

Problem 5: Solve the quadratic equation 16x² - 24x = 0

Solution:

The given equation may be written as 8x(2x - 3) = 0

This gives x = 0 or x = 3/2
x = 0, 3/2 are the required solutions.

Problem 6: Solve: 25x² - 30x + 9 = 0

Solution:

 25x² - 30x + 9 = 0 is equivalent to (5x)² - 2(5x) × 3 + (3)² = 0

or (5x - 3)² = 0

This gives x = 3/5, 3/5 or simply x = 3/5 as the required solution.

Problem 7: For what value of k, (4 - k)x² + (2k + 4)x + (8k + 1) is a perfect square?

Solution:

The given equation is a perfect square, if its discriminant is zero i.e., (2k + 4)² - 4(4 - k)(8k + 1) = 0

⟹ 4(k + 2)² - 4(4 - k)(8k + 1) = 0

⟹ 4[4(k + 2)² - (4 - k)(8k + 1)] = 0

⟹ [(k² + 4k + 4) - (-8k² + 31k + 4)] = 0

⟹ 9k² - 27k = 0

⟹ 9k(k - 3) = 0

⟹ k = 0 or k = 3

Hence, the given equation is a perfect square, if k = 0 or k = 3.

Problem 8: Out of a group of swans, 7/2 times the square root of the total number are playing on the shore of a tank. The two remaining ones are playing in deep water. What is the total number of swans?

Solution:

Let us denote the number of swans by x.

Then, the number of swans playing on the shore of the tank = (7/2)√x

There are two remaining swans.

Therefore, x = (7/2)√x + 2

or x - 2 = (7/2)√x

or (x - 2)² = (7/2)²x

or 4(x² - 4x + 4) = 49x

or 4x² - 65x + 16 = 0

or 4x² - 64x - x + 16 = 0

or 4x(x - 16) - 1(x - 16) = 0

or (x - 16)(4x - 1) = 0

This gives x = 16 or x = 1/4

We reject x = 1/4 and take x = 16.

Hence, the total number of swans is 16.

Problem 9: The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.

Solution:

Let the length of the shorter side be x cm. Then, the length of the longer side = (x + 5) cm.

Since the triangle is right-angled, the sum of the squares of the sides must be equal to the square of the hypotenuse (Pythagoras Theorem).

x² + (x + 5)² = 25²

or x² + x² + 10x + 25 = 625

or 2x² + 10x - 600 = 0

or x² + 5x - 300 = 0

or (x + 20)(x - 15) = 0

This gives x = 15 or x = -20

We reject x = -20 and take x = 15.

Thus, length of shorter side = 15 cm.

Length of longer side = (15 + 5) cm, i.e., 20 cm.

Problem 10: Swati can row her boat at a speed of 5 km/h in still water. If it takes her 1 hour more to row the boat 5.25 km upstream than to return downstream, find the speed of the stream.

Solution:

Let the speed of the stream be x km/h

∴ Speed of the boat in upstream = (5 - x) km/h

Speed of the boat in downstream = (5 + x) km/h

Time, say t₁ (in hours), for going 5.25 km upstream = 5.25/(5 - x)

Time, say t₂ (in hours), for returning 5.25 km downstream = 5.25/(5 + x)

Obviously t₁ > t₂

Therefore, according to the given condition of the problem, t₁ = t₂ + 1

i.e., 5.25/(5 - x) = 5.25/(5 + x) + 1

or 21/4 × (1/(5 - x) - 1/(5 + x)) = 1

or 21 × (5 + x - 5 + x)/4(25 - x²) = 1

or 42x = 100 - 4x²

or 4x² + 42x - 100 = 0

or 2x² + 21x - 50 = 0

or (2x + 25)(x - 2) = 0

This gives x = 2, since we reject x = -25/2.

Thus, the speed of the stream is 2 km/h.

Illustration 1: Determine whether the given values of x are solutions of the given equation or not:
x² + 2√3x - 9 = 0; x = √3, x = -3√3

Solution: Consider p(x) = x² + 2√3x - 9 ...(i)

Substituting x = √3 in (i):

p(√3) = (√3)² + 2√3(√3) - 9 = 3 + 6 - 9 = 0

∴ x = √3 is a solution of the given quadratic equation.

Substituting x = -3√3 in (i) we get:

p(-3√3) = (-3√3)² + 2√3(-3√3) - 9 = 27 - 18 - 9 = 0 = RHS

∴ x = -3√3 is a solution of the given quadratic equation.

Illustration 2: If x = 2 and x = 3 are roots of 3x² - 2px + 2q = 0, find the values of p and q.

Solution: Since x = 2 is a root of the given equation,

∴ we have 3(2)² - 2p(2) + 2q = 0

⟹ 12 - 4p + 2q = 0

⟹ 4p - 2q = 12 ...(i)

Since x = 3 is a root of the given equation,

∴ we have 3(3)² - 2p(3) + 2q = 0

⟹ 27 - 6p + 2q = 0

⟹ 6p - 2q = 27 ...(ii)

Subtracting equation (i) from equation (ii), we get

6p - 2q - 4p + 2q = 27 - 12

or 2p = 15

⟹ p = 15/2

Substituting p = 15/2 in equation (i), we get

4(15/2) - 2q = 12

⟹ 2q = 30 - 12

⟹ 2q = 18

⟹ q = 9
Hence p = 15/2 and q = 9.

Illustration 3: Solve: a²b²x² + b²x - a²x - 1 = 0

Solution: a²b²x² + b²x - a²x - 1 = 0

⟹ b²x(a²x + 1) - 1(a²x + 1) = 0

⟹ (a²x + 1)(b²x - 1) = 0

Either (a²x + 1) = 0 or (b²x - 1) = 0

⟹ a²x = -1 or b²x = 1

⟹ x = -1/a² or x = 1/b²
Hence, x = -1/a², 1/b² are the required solutions.