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
pm x pn = p m + n
{pm} / {pn} = p m – n
(pm)n = pmn
p-m = 1/pm
p1 = p
p0 = 1