Question

How do you solve the equation ${x^2} - 4x + 1 = 0$ by completing the square?

Answer 1

Hint:
Whenever we need to factorise any quadratic equation we need to add and subtract the value of ${\left( {\dfrac{{{\text{coefficient of }}x}}{2}} \right)^2}$ in the given quadratic equation. Now we will get the equation in form of the ${\left( {a + b} \right)^2} \pm {\text{c or }}{\left( {a - b} \right)^2} \pm {\text{c }}$ and now we can simplify it further as much as we can.

Complete step by step solution:
Here we are given the quadratic equation which we need to factorise. So we must know that any equation of degree $2$ is known as the quadratic equation. In order to find its factor we need to add and subtract the value of ${\left( {\dfrac{{{\text{coefficient of }}x}}{2}} \right)^2}$ in the given quadratic equation.
So here we can see in the above equation which is ${x^2} - 4x + 1 = 0$ that the coefficient of $x{\text{ is }} - 4$.
So we need to add and subtract the value ${\left( {\dfrac{{ - 4}}{2}} \right)^2} = \dfrac{{16}}{4} = 4$ in the above quadratic equation which is given. Hence we will get:
${x^2} - 4x + 1 + 4 - 4 = 0$
Now we can write it in the form:
$
  {x^2} - 4x + 4 + 1 - 4 = 0 \\
  \left( {{x^2} - 4x + 4} \right) - 3 - - - - - - (1) \\
 $
Now we know that we can write the term in the first bracket of the equation (1) as:
$\left( {{x^2} - 4x + 4} \right) = {\left( {x - 2} \right)^2}$
Now we can substitute this calculated value in the equation (1) and we will get:
${\left( {x - 2} \right)^2} - 3$
Now we can write it as:
${\left( {x - 2} \right)^2} - 3$
${\left( {x - 2} \right)^2} - {\left( {\sqrt 3 } \right)^2}$
Now we know that:
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
So we can apply it over here and get:
${\left( {x - 2} \right)^2} - {\left( {\sqrt 3 } \right)^2}$
$\left( {x - 2 + \sqrt 3 } \right)\left( {x - 2 - \sqrt 3 } \right) = 0$
On simplifying it we will get:
$
  \left( {x - (2 - \sqrt 3 )} \right)\left( {x - (2 + \sqrt 3 )} \right) = 0 \\
  x = 2 + \sqrt 3 {\text{ or 2}} - \sqrt 3 \\
 $

Hence we get the quadratic equation result as $2 + \sqrt 3 {\text{ or }}2 - \sqrt 3 $

Note:
If we were not told to find by completing the square we would have done by using the formula where roots of the quadratic equation of the general form $a{x^2} + bx + c = 0$ are given by:
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$