Class X – Maths Formulas – Quadratic Equations

Quadratic Equations

1. Quadratic Polynomial	P(𝑥) = a 𝑥2 + b 𝑥 + c, where a ≠ 0
2. Quadratic Equation	a 𝑥2 + b 𝑥 + c, where a ≠ 0
3. Solution or root of the Quadratic Equation	A real number α is called the root or solution of the quadratic equation if a α2 + b α + c = 0
4. Zeroes of the polynomial P(𝑥)	The root of the quadratic equation are called zeroes.
5. Maximum roots of quadratic equations	A polynomial of degree 2 can have max 2 roots.
6. Condition for real roots	A quadratic equation has real roots if b2 – 4ac > 0

Solving Quadratic Equation

Factorization

This method we factorize the equation by splitting the middle term b. In, a 𝑥2 + b 𝑥 + c = 0 Example: 6 𝑥2 – 𝑥 – 2 = 0

Steps: 1. First we need to multiply the coefficient of a and c. In this case: 6 x (-2) = -12 2. Splitting the middle term so that multiplication is -12 and the difference is the coefficient b 6 𝑥2 + 3 𝑥 – 4 𝑥 – 2 = 0 3 𝑥 (2 𝑥 + 1) – 2(2 𝑥 + 1) = 0 (3 𝑥 – 2) (2 𝑥 + 1) = 0 3. Roots of the equation can be found by equating the factors to zero 3 𝑥 – 2 = 0 => 𝑥 = 2/3 2 𝑥 + 1 = 0 => 𝑥 = -1/2

Square Method

In this method we create square on LHS and RHS and then find the value.

a 𝑥2 + b 𝑥 + c = 0 1) 𝑥2 + (b/a) 𝑥 + (c/a) = 0 2) (𝑥 + b/2a)2 – (b/2a)2 + (c/a) = 0 3) (𝑥 + b/2a)2 = (b2 – 4ac)/4a2 4) 𝑥 = [-b ± √(b2 – 4ac)]/2a

Example: 𝑥2 + 4 𝑥 – 5 = 0 1) (𝑥 + 2)2 – 4 – 5 = 0 2) (𝑥 + 2)2 = 9 3) Roots of the equation can be found using square root on both the sides 𝑥 + 2 = -3 => 𝑥 = -5 𝑥 + 2 = 3 => 𝑥 = 1

Quadratic Method

For quadratic equation, a 𝑥2 + b 𝑥 + c = 0 Roots are given by 𝑥 = [-b + √ (b2 – 4ac)]/2a 𝑥 = [-b – √ (b2 – 4ac)]/2a For b2 – 4 ac > 0, Quadratic equation has two real roots of different value For b2 – 4 ac = 0, Quadratic equation has one real root. For b2 – 4 ac < 0, Quadratic equation has no real root.

Nature of roots of Quadratic Equation:

1. b2 – 4 ac > 0	Two distinct real roots
2. b2 – 4 ac = 0	One real root
3. b2 – 4 ac < 0	No real roots

Algebraic Formulas

(a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

(a + b) (a – b) = a² – b²

(𝑥 + a) (𝑥 + b) = 𝑥² + (a + b) 𝑥 + ab

(𝑥 + a) (𝑥 – b) = 𝑥² + (a – b) 𝑥 – ab

(𝑥 – a) (𝑥 + b) = 𝑥² + (b – a) 𝑥 – ab

(𝑥 – a) (𝑥 – b) = 𝑥² – (a + b) 𝑥 + ab

(a + b)³ = a³ + b³ + 3ab (a + b)

(a – b)³ = a³ – b³ – 3ab (a – b)

(𝑥 + 𝑦 + z)² = 𝑥² + 𝑦² + z² + 2𝑥𝑦 + 2𝑦z + 2z𝑥

(𝑥 + 𝑦 – z)² = 𝑥² + 𝑦² + z² + 2𝑥𝑦 – 2𝑦z – 2z𝑥

(𝑥 – 𝑦 + z)² = 𝑥² + 𝑦² + z² – 2𝑥𝑦 – 2𝑦z + 2z𝑥

(𝑥 – 𝑦 – z)² = 𝑥² + 𝑦² + z² – 2𝑥𝑦 + 2𝑦z – 2z𝑥

𝑥³ + 𝑦³ + z³ – 3𝑥𝑦z = (𝑥 + 𝑦 + z) (𝑥² + 𝑦² + z² – 𝑥𝑦 – 𝑦z – z𝑥)

𝑥² + 𝑦² = ½ [(𝑥 + 𝑦)² + (𝑥 – 𝑦)²]

 (𝑥 + a) (𝑥 + b) (𝑥 + c) = 𝑥³ + (a + b + c) 𝑥² + (ab + bc + ca) 𝑥 + abc

 𝑥³ + 𝑦³ = (𝑥 + 𝑦) (𝑥² – 𝑥𝑦 + 𝑦²)  

𝑥³ – 𝑦³ = (𝑥 – 𝑦) (𝑥² + 𝑥𝑦 + 𝑦²)  

𝑥² + 𝑦² + z² – 𝑥𝑦 – 𝑦z – z𝑥 = ½ [((𝑥 – 𝑦)² + (𝑦 – z)² + (z –  𝑥)²]

Basic formulas for powers

p^m x pⁿ  = p ^{m + n}

{p^m} / {pⁿ} = p ^{m – n}

(p^m)ⁿ = p^mn

p^-m = 1/p^m

p¹ = p

p⁰ = 1