The term 'quadratic' comes from a Latin word 'quadratus', which means 'square'. So, any equation having two as the maximum value of power, can be called a 'quadratic equation'. In the following lines, I will be defining some important terms before getting down to solving quadratic equations by factorization method using simple examples. Definitions of a Quadratic Equation Quadratic Polynomial A polynomial of a second degree is called a quadratic polynomial. The general form of a quadratic polynomial is ax² + bx + c, where a, b, c are real numbers, a ≠ 0 and x is a variable. Example x² + 2x + 1; 3x² + √6x Quadratic Equation An equation p(x) = 0, where p(x) is a quadratic polynomial, is called a quadratic equation. The general form of a quadratic equation is, ax² + bx + c = 0 where a, b, c are real numbers, a ≠ 0 and x is a variable. Example x² - 6x + 2 = 0 Roots of a Quadratic Equation There are three different methods to find the roots of any quadratic equation. We can find the roots using factorization method, completing the square method and by using a formula. Among all these methods, factorization is a very easy method. The roots of a quadratic equation are the values of 'x', which should satisfy the given equation. An important point to be noted is that, a quadratic equation cannot have more than two distinct roots. The roots always exists in a pair. Before starting to solve the quadratic equation, follow the steps below.

• Consider the general form of a quadratic equation i.e., ax² + bx + c = 0.
• Factorize the term 'ac' such that the sum of the factors is equal to b.

With this, let us start solving the problems by method of factorization by splitting the middle term. Solving Problems by Factorization Method

(i)

x2 - 5x + 6 = 0

Solution a = 1, b = -5 and c = 6; a × c = 1 × 6 = 6; The factors of 6 whose sum is equal to -5 is -3 and -2; -5x can be replaced with -3x and -2x; x2 - 3x -2x + 6 = 0; x(x-3) - 2(x-3) = 0; Now, collect the common term in the bracket (x-3) and make the equation in the following way, (x-3)(x-2) = 0; If the product of a and b is zero, i.e., ab = 0, then either a = 0 or b = 0; ∴ x-2 = 0 or x-3 = 0; So, x = 2 or x = 3 Thus, the roots of the quadratic equation x2 - 5x + 6 = 0 are 3 and 2.

(ii)	2x2 + x - 3 = 0
Solution a = 2, b = 1 and c = -3; a × c = -6; The factors of -6 whose sum is equal to 1 is 3 and -2; 2x2 + 3x -2x - 3 = 0; 2x2 - 2x + 3x - 3 = 0; 2x (x-1) + 3 (x-1) = 0: (2x+3)(x-1) = 0; ∴ 2x+3 = 0 or x-1 = 0; x = 1 or x = - 3/2 Thus, the roots of the quadratic equation 2x2 + x - 3 = 0 are 2 and -3/2.

(iii)	x (x + 7) = 0
Solution This is a simple problem but there is a common error which people make, by multiplying the bracket term (i.e) x+7 with the x and making it a quadratic equation and to factorize it. But the logic is that, it is already factored! (x+0)(x+7) = 0; ∴ (x+0) = 0 or (x+7) = 0; Thus x = 0 or x = -7 Thus, the roots of the quadratic equation x (x + 7) = 0 are 0 and -7.

(iv)	x2 - 25 = 0
Solution x2 - 52 = 0; The quadratic equation is of the form a2 - b2 = (a+b)(a-b); So, (x+5)(x-5) = 0; ∴ (x+5) = 0 or (x-5) = 0; Thus, x = -5 or x = 5

How to Check the Quadratic Equation If you are a beginner, it's always better to check the results for your confirmation.

Substitute the value of the roots in the given quadratic equation of the first problem. The roots which we obtained for the above question are 3 and 2. Now, substitute the value of x in x2 - 5x + 6 = 0; for x = 3; x2 - 5x + 6 = 0; L.H.S (Left Hand Side)= 32 - 5(3) + 6; = 9 - 15 + 6; = 0 = R.H.S (Right Hand Side) proved. By doing this, you can be confident that your solution is right.

When you are clear with the basics of solving quadratic equation by factoring, then solving it will be the easiest one in algebraic mathematics. At the beginning stage, practice the problems with the step by step process, which I have explained above. After adequate practice, you will be able to solve any quadratic equation, in no time.