Question

How do you solve by completing the square $ {x^2} + 8x + 1 = 0 $ ?

Answer 1

Hint: To determine the solution to the above equation using completing the square method, first divide the equation by the coefficient of $ {x^2} $ to convert into $ {x^2} + bx $ form and add $ {\left( {\dfrac{b}{2}} \right)^2} $
 $ {\left( {\dfrac{b}{2}} \right)^2} $ to make $ {x^2} + bx + {\left( {\dfrac{b}{2}} \right)^2} $ which is a perfect square of $ {\left( {x + \dfrac{b}{2}} \right)^2} $ .Transposing and combining constant terms on right hand side and taking square root on both sides will lead to your required result.

Complete step-by-step solution:
We are given a quadratic equation $ {x^2} + 8x + 1 = 0 $
First divide the whole equation with the coefficient of $ {x^2} $ to obtain the form $ {x^2} + bx $ in the equation only if the coefficient is not equal to 1.
Since in our case the coefficient is already equal to one, so no need to divide.
 $ {x^2} + 8x + 1 = 0 $
We are going to rearrange $ {x^2} + bx + e $ in order to complete it into a square by adding $ {\left( {\dfrac{b}{2}} \right)^2} $ to make $ {x^2} + bx + {\left( {\dfrac{b}{2}} \right)^2} $ which is a perfect square of $ {\left( {x + \dfrac{b}{2}} \right)^2} $ where b is the coefficient of x .
\[{\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{8}{2}} \right)^2} = {\left( 4 \right)^2} = 16\]
Therefore we have to add $ 16 $ in the equation to make it a square but we can’t simply add $ 16 $ on LHS as this will disturb the balance, so we will add $ 16 $ on both sides of the equation. Our equation will now become .
\[ \Rightarrow {x^2} + 8x + 16 + 1 = + 16\]
\[{x^2} + 8x + 16\] can be written as $ {\left( {x + 4} \right)^2} $
 $
   \Rightarrow {\left( {x + 4} \right)^2} + 1 = 16 \\
   \Rightarrow {\left( {x + 1} \right)^2} = 16 - 1 \\
   \Rightarrow {\left( {x + 1} \right)^2} = 15 \\
   \Rightarrow x + 1 = \pm \sqrt {15} \\
   \Rightarrow x = - 1 \pm \sqrt {15} \\
  $
Therefore, the solution to $ {x^2} + 8x + 1 = 0 $ is equal to $ x = - 1 - \sqrt {15} , - 1 + \sqrt {15} $

Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $ a{x^2} + bx + c $ where $ x $ is the unknown variable and a,b,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $ D = {b^2} - 4ac $
Using Discriminant, we can find out the nature of the roots
-If D is equal to zero, then both of the roots will be the same and real.
-If D is a positive number then, both of the roots are real solutions.
-If D is a negative number, then the root are the pair of complex solutions

Note:
1. One must be careful while calculating the answer as calculation error may come.
2. Don’t forget to compare the given quadratic equation with the standard one every time.
3.Completing the square may not be an efficient method in every case of solving quadratic equations.