Converting Quadratic Equations

INTRODUCTION

Any equation which has been written in the form  ax 2+bx + c = 0 is called as quadratic equations. In quadratic equations the numbers a, b, and c are real numbers and a ? zero. If p(x) is an equation and which is the polynomial having the degree of 2 and in the form of p(x)= 0 , then the equation is  quadratic.

Steps for Converting Quadratic Equations:

The General form of quadratic equation is: y=ax2+bx+c

The Vertex form   is: y= a(x-h) 2 +k.

The following steps are used for converting quadratic equations,

Step 1: The initial step is to factor out the leading co-efficient from the first two terms of the given equation.

Step 2:  Complete the square to which we have factored out (First two terms).

Step 3: Balance the constant term we get after completing the square, by multiplying it with the co-efficient.

Step 4: Add the opposite to the constant term of given equation.

Algebra is widely used in day to day activities watch out for my forthcoming posts on help for algebra 2 and simplify any algebraic expression. I am sure they will be helpful.

Example Problems on Converting Quadratic Equations :
Converting quadratic equations Example 1:

Find the vertex form of the given quadratic equation y = 4x² - 16x + 6

Solution:

Step 1: Factor out the leading co-efficient from the first two terms of the given equation.

Y= 4 (x2-4x) + 6

Step 2:  Complete the square to which we have factored out (First two terms).

Y= 4 (x2-4x) + 6

(x2-4x) = (x-2)2   { (a-b)2 = a2-2ab+b2  , here a= x, b=2 }

= x2-(2)(x)(2) +22

y = x2-4x+4

Step 3: Balance the constant term 4, by multiplying it with the co-efficient 4.

y  = 4(x2-4x+4)  +6 +4(4)

Step 4: Add the opposite to the constant term of given equation.

y= 4(x-2)2   +6-16

Vertex form is   :  y= 4(x-2)2   -10

Vertex            :   (h=2, k= -10).

Converting quadratic equationsExample 2:

Convert the quadratic equation y = -5x2+30 into vertex form.

Solution:

Step 1: Factor out the leading co-efficient from the first two terms of the given equation.

y = -5x2+30

y= -5(x2-6x)

Step 2:  Complete the square to which we have factored out

y= -5(x2-6x)

(x2-6x) = (x-3)2   { (a-b)2 = a2-2ab+b2   , Here a= x, b= 3}

= x2-(2)(x)(3)+ 32

y= x2-6x+9

Step 3: Balance the constant term 9, by multiplying it with the co-efficient -5

y= -5(x2-6x+9) + (-5)(9)

y  = -5(x2-6x+9)-45

Step 4: Add the opposite to the constant term of given equation.

Here there is no constant term .So just change the sign of -45.                                 

y= -5(x-3)2 +45

Vertex form is   :  y= -5(x-3)2 +45

Vertex      :  ( h=3, k=45)