Question

How do you solve by completing the square $ 2{x^2} + 4x + 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} $ .

Complete step-by-step answer:
We are given a quadratic equation $ 2{x^2} + 4x + 1 = 0 $
First divide the whole equation with the coefficient of $ {x^2} $ as to obtain the form $ {x^2} + bx $ in the equation
Dividing the equation by 2 on both the sides
 $
   \Rightarrow \dfrac{1}{2}\left( {2{x^2} + 4x + 1} \right) = 0 \times \dfrac{1}{2} \\
   \Rightarrow {x^2} + 2x + \dfrac{1}{2} = 0 \;
  $
We are going to rearrange $ {x^2} + bx + e $ in order to complete it in to 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{2}{2}} \right)^2} = 1 $
Therefore we have to add 1 in the equation to make it a square but we can’t simply add 1 on LHS as this will disturb the balance, so we will add 1 on both sides of the equation. Our equation will now become .
 $ \Rightarrow {x^2} + 2x + 1 + \dfrac{1}{2} = 1 $
 $ {x^2} + 2x + 1 $ can be written as $ {\left( {x + 1} \right)^2} $
 $
   \Rightarrow {\left( {x + 1} \right)^2} + \dfrac{1}{2} - 1 = 0 \\
   \Rightarrow {\left( {x + 1} \right)^2} - \dfrac{1}{2} = 0 \\
  $
 $ \dfrac{1}{2} $ can be written as \[\dfrac{1}{{{{\left( {\sqrt 2 } \right)}^2}}} = {\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2}\]
 $ \Rightarrow {\left( {x + 1} \right)^2} - {\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2} = 0 $
Using the identity $ ({A^2} - {B^2}) = (A + B)(A - B) $ Where A is $ x + 1 $ and B as $ \dfrac{{\sqrt 2 }}{2} $
 $
   \Rightarrow {\left( {x + 1} \right)^2} - {\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2} = 0 \\
   \Rightarrow \left( {x + 1 + \dfrac{{\sqrt 2 }}{2}} \right)\left( {x + 1 - \dfrac{{\sqrt 2 }}{2}} \right) = 0 \\
  x + 1 + \dfrac{{\sqrt 2 }}{2} = 0 \\
   \Rightarrow x = - 1 - \dfrac{{\sqrt 2 }}{2} \\
  x + 1 - \dfrac{{\sqrt 2 }}{2} = 0 \\
   \Rightarrow x = - 1 + \dfrac{{\sqrt 2 }}{2} \;
  $
Therefore, the solution to $ 2{x^2} + 4x + 1 = 0 $ is equal to $ x = - 1 - \dfrac{{\sqrt 2 }}{2}, - 1 + \dfrac{{\sqrt 2 }}{2} $
So, the correct answer is “$ x = - 1 - \dfrac{{\sqrt 2 }}{2}, - 1 + \dfrac{{\sqrt 2 }}{2} $
”.

Note: 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