Complete List Of All Important Formulas

Certainly! Here’s a professional, well-organized list of important formulas for Class 10 Mathematics, from Chapter 1 to Chapter 5, with brief introductions, explanations, and examples for each chapter: Chapter 1: Real Numbers Introduction: In this chapter, we explore the concept of real numbers, their types, and operations with a focus on rational and irrational numbers. Important Formulas: Euclid’s Division Lemma Formula: 𝑎 = 𝑏 𝑞 + 𝑟 a=bq+r where: 𝑎 a and 𝑏 b are integers, 𝑞 q is the quotient, and 𝑟 r is the remainder. Explanation: This lemma helps to find the greatest common divisor (GCD) of two numbers. Example: Question: Divide 65 by 24 and find the quotient and remainder. Solution: 65 = 24 × 2 + 17 65=24×2+17, so quotient = 2, remainder = 17. Fundamental Theorem of Arithmetic Formula: Every integer greater than 1 can be written as a product of prime numbers in a unique way, except for the order of factors. Explanation: This theorem forms the basis of prime factorization. Example: Question: Find the prime factorization of 180. Solution: 180 = 2 2 × 3 2 × 5 180=2 2 ×3 2 ×5 Summary: In this chapter, we learned Euclid’s Division Lemma and the Fundamental Theorem of Arithmetic for prime factorization. Chapter 2: Polynomials Introduction: This chapter deals with polynomials, their degrees, types, and operations. Important Formulas: Standard Form of a Polynomial Formula: 𝑃 ( 𝑥 ) = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛 − 1 𝑥 𝑛 − 1 + ⋯ + 𝑎 1 𝑥 + 𝑎 0 P(x)=a n x n +a n−1 x n−1 +⋯+a 1 x+a 0 Explanation: A polynomial is an expression consisting of variables raised to non-negative integer powers with coefficients. Example: Question: Write the polynomial for 𝑥 3 − 4 𝑥 2 + 3 𝑥 − 7 x 3 −4x 2 +3x−7 in standard form. Solution: The polynomial is already in standard form: 𝑥 3 − 4 𝑥 2 + 3 𝑥 − 7 x 3 −4x 2 +3x−7. Zero of a Polynomial Formula: If 𝑃 ( 𝑥 ) = 0 P(x)=0, then 𝑥 x is a zero of the polynomial. Explanation: Zeros are the values of 𝑥 x that make the polynomial equal to zero. Example: Question: Find the zeros of 𝑃 ( 𝑥 ) = 𝑥 2 − 5 𝑥 + 6 P(x)=x 2 −5x+6. Solution: 𝑥 2 − 5 𝑥 + 6 = 0 ⇒ ( 𝑥 − 2 ) ( 𝑥 − 3 ) = 0 x 2 −5x+6=0⇒(x−2)(x−3)=0, so the zeros are 𝑥 = 2 x=2 and 𝑥 = 3 x=3. Summary: In this chapter, we learned the standard form of polynomials and the concept of zeros. Chapter 3: Pair of Linear Equations in Two Variables Introduction: This chapter focuses on solving systems of linear equations using various methods like substitution, elimination, and graphical representation. Important Formulas: General Form of a Linear Equation Formula: 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 = 0 ax+by+c=0 Explanation: A linear equation is an equation involving two variables, where the highest power of the variables is 1. Example: Question: Write the equation of a line with slope 2 and y-intercept -3. Solution: The equation is 𝑦 = 2 𝑥 − 3 y=2x−3, or 2 𝑥 − 𝑦 − 3 = 0 2x−y−3=0 in general form. Substitution Method Formula: Solve one equation for one variable and substitute it in the other. Explanation: This method helps to solve a pair of linear equations by substitution. Example: Question: Solve 𝑥 + 𝑦 = 6 x+y=6 and 𝑥 − 𝑦 = 2 x−y=2 using substitution. Solution: From 𝑥 + 𝑦 = 6 x+y=6, 𝑥 = 6 − 𝑦 x=6−y. Substituting into 𝑥 − 𝑦 = 2 x−y=2, ( 6 − 𝑦 ) − 𝑦 = 2 ⇒ 6 − 2 𝑦 = 2 ⇒ 𝑦 = 2 (6−y)−y=2⇒6−2y=2⇒y=2, and then 𝑥 = 6 − 2 = 4 x=6−2=4. Summary: We explored the general form of linear equations and the substitution method to solve them. Chapter 4: Quadratic Equations Introduction: This chapter deals with solving quadratic equations using methods such as factoring, completing the square, and using the quadratic formula. Important Formulas: Standard Form of a Quadratic Equation Formula: 𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0 ax 2 +bx+c=0 Explanation: A quadratic equation is a second-degree polynomial equation in one variable. Example: Question: Solve 𝑥 2 − 5 𝑥 + 6 = 0 x 2 −5x+6=0. Solution: Factorizing, ( 𝑥 − 2 ) ( 𝑥 − 3 ) = 0 (x−2)(x−3)=0, so 𝑥 = 2 x=2 and 𝑥 = 3 x=3. Quadratic Formula Formula: 𝑥 = − 𝑏 ± 𝑏 2 − 4 𝑎 𝑐 2 𝑎 x= 2a −b± b 2 −4ac Explanation: This formula provides the roots of any quadratic equation 𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0 ax 2 +bx+c=0. Example: Question: Solve 2 𝑥 2 − 4 𝑥 − 6 = 0 2x 2 −4x−6=0 using the quadratic formula. Solution: Using the formula: 𝑥 = − ( − 4 ) ± ( − 4 ) 2 − 4 ( 2 ) ( − 6 ) 2 ( 2 ) x= 2(2) −(−4)± (−4) 2 −4(2)(−6) 𝑥 = 4 ± 16 + 48 4 = 4 ± 64 4 = 4 ± 8 4 x= 4 4± 16+48 = 4 4± 64 = 4 4±8 So, 𝑥 = 3 x=3 or 𝑥 = − 1 x=−1. Summary: We learned the standard form of a quadratic equation and how to solve it using factoring and the quadratic formula. Chapter 5: Arithmetic Progressions (AP) Introduction: In this chapter, we explore the concept of Arithmetic Progressions (AP), a sequence of numbers in which the difference between consecutive terms is constant. Important Formulas: nth Term of an AP Formula: 𝑎 𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 a n =a+(n−1)d where: 𝑎 𝑛 a n is the nth term, 𝑎 a is the first term, 𝑑 d is the common difference, and 𝑛 n is the term number. Explanation: This formula helps to find any term in an AP. Example: Question: Find the 10th term of the AP 2 , 5 , 8 , 11 , … 2,5,8,11,…. Solution: Here, 𝑎 = 2 a=2, 𝑑 = 3 d=3. 𝑎 10 = 2 + ( 10 − 1 ) × 3 = 2 + 27 = 29 a 10 =2+(10−1)×3=2+27=29. Sum of First n Terms of an AP Formula: 𝑆 𝑛 = 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ] S n = 2 n [2a+(n−1)d] Explanation: This formula helps to calculate the sum of the first 𝑛 n terms in an AP. Example: Question: Find the sum of the first 5 terms of the AP 3 , 7 , 11 , 15 , … 3,7,11,15,…. Solution: Here, 𝑎 = 3 a=3, 𝑑 = 4 d=4, and 𝑛 = 5 n=5. 𝑆 5 = 5 2 [ 2 ( 3 ) + ( 5 − 1 ) × 4 ] = 5 2 [ 6 + 16 ] = 5 2 × 22 = 55 S 5 = 2 5 [2(3)+(5−1)×4]= 2 5 [6+16]= 2 5 ×22=55. Summary: We explored the nth term formula and the sum of the first 𝑛 n terms of an AP. Conclusion: These are the key formulas from Chapters 1 to 5 of Class 10 Mathematics. Each chapter introduces fundamental concepts, and the formulas provide tools to solve various mathematical problems.