Solving Quadratic Equations

An equation of the form ax2+bx+c=y where ‘a’ is not equal to zero; a, b are the coefficients and c is a constant is called a Quadratic Equation.

There are different methods used in solving these equations. They are, factorization method, grouping factors method, completing square method, using the quadratic formula if the equation does not factor and graphing method.

Solving Quadratic Equations Examples are as follows: Solve 2x2+5x-3=0
Let us solve this using the factorization method. Here first find the product of the coefficient of x2 term and the constant term, (2)(-3)=-6. Now find the factors of -6 such that the sum of the factors is equal to the coefficient of the middle term(x). The list of factors of -6 can be given as, -2x3, -3x2, -1x6, 6x-1.


Now select the appropriate set of factors whose sum would be equal to 5.
That would be 6, -1 as 6-1=5. Let us now re-write the given equation by splitting the middle term into the factors, which gives
2x2+6x –x +3=0. Now re-grouping the terms we get, 2x(x+3)-1(x+3). Taking (x+3) results in, (x+3)(2x-1). So the factors of the given equation are (2x-1) and (x+3). Each of these factors is equated to zero and solved to arrive to the solution set of the equation. 2x-1=0; x=1/2 and x+3=0; x=-3. So, the solution set is x=1/2, x=-3.

The above method can be used only when the given equation can be factored. There are some cases where the equation cannot be factored. In such cases the Quadratic Formula Help solving Quadratic Equations. The standard Quadratic Equation being ax2+bx+c=0, the quadratic formula is given by,
x= -bsqrt[b2-4ac] divided by 2a.

There would be two solutions, one with positive square root and other with negative square root. Let us consider an example problem, solve x2+3x-5=0. Here first the given equation is compared with the standard form to get the values of a,b and c respectively.

By comparing the terms to the standard form ax2+bx+c=0 we get, a=1, b=3 and c=-5. Now these values are substituted in the quadratic formula, x= [-bsqrt(b2-4ac)]/2a which gives x= [-3sqrt(9 –(4)(1)(-5))]/2(1)
= -3sqrt[9+20]/2 = [-3]/2. The solutions for the given equation are x=[3+29]/2 and x=[3-29]/2

Thus using the quadratic formula it becomes easy to solve equations which cannot be factored. This formula can also be used to an equation that can be factored too. Some of the Quadratic Equation Practice Problems are:
Solve 2x2-4x-3=0
Solve x2-14x+45=0
Solve -3x2+x+5=0
Solve 4x-5-3x2=0
Solve x2-x-6=0

Algebra is widely used in day to day activities watch out for my forthcoming posts on Finding Volume of a Cube and cbse 11th sample paper. I am sure they will be helpful.