Question

Find the roots for x×x+6x+5 by completing square method

Answer 1

Given the equation x² + 6x + 5 = 0, we will follow the following steps:

The first step is to analyze the number that is multiplying the term x².
• If the number is different from 1 we will divide both sides of the equation by this number;
• If the number equals 1 we do not need to make any changes to the equation and move on to the next step.
 
The second step is to add to both sides of the equation the square of the half of the number that is multiplying the term "x" of our equation. The number that is multiplying the term "x" of the equation is equal to 6. To find the square of the half of that number we just divide it by 2 and then raise the result squared. Nothing too hard to do, see:
 
6/2 = 3 => 3² = 9
 
We found that the square of the half of 3 is 9, so we'll add this number to both sides of our equation as the second step orders. Here's how our equation will look:
 
x² + 6x + 5 + 9 = 0 + 9
 
The most interesting part comes now. When we add the square of the half of the term that multiplies the "x" to both members of our equation we transform the first member of it into a perfect square trinomial. Look:
 
(x² + 6x + 9) + 5 = 9
 
The term in parentheses is a perfect square trinomial that can be expressed as follows:
 
(x² + 6x + 9) → (x + 3)²
 
Substituting this into the equation we have:
(x² + 6x + 9) + 5 = 9
(x + 3)² + 5 = 9
(x + 3)² = 9 - 5
√(x + 3)² = ±√4  
x + 3 = ±√4
x + 3 = ±2

x + 3 = 2 or x + 3 = - 2

x = 2 - 3 or x = - 2 - 3

$\boxed{x = -1}\: or\: \boxed{x = - 5}$
 
We find the roots of our equation without using the famous Bhaskara formula. These steps are applicable to any 2nd degree equation of the type: ax² + bx + c = 0
with "a" different from zero and "b" and "c" belonging to the set of real numbers.