Question

How do you solve ${x^2} - 3x + 2 = 0$by completing the square?

Answer 1

Hint: Completing the square is the method which represents the quadratic equation as the combination of the quadrilateral used to form the square and it is the basis of the method discovers the special value which when added to both the sides of the quadratic which creates the perfect square trinomial. Here we will take the given expression and check for the perfect square or the value to be added.

Complete step by step answer:
Take the given expression: ${x^2} - 3x + 2 = 0$
Move the term without any variable that is constant on the opposite side. When you move any term from one side to another the sign of the term also changes.
${x^2} - 3x = - 2$ …. (A)
Now to create the trinomial square on the left hand side of the equation find the value which is equal to the square of half of the “b”
${\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{3}{2}} \right)^2}$
Add above term on both the sides of the equation (A)
${x^2} - 3x + {\left( {\dfrac{3}{2}} \right)^2} = - 2 + {\left( {\dfrac{3}{2}} \right)^2}$
Simplify the above equation –
${x^2} - 3x + \left( {\dfrac{9}{4}} \right) = - 2 + \left( {\dfrac{9}{4}} \right)$
Take LCM (least common multiple) on the right hand side of the equation and simplify it.
${x^2} - 3x + \left( {\dfrac{9}{4}} \right) = \left( {\dfrac{1}{4}} \right)$
Now, the left hand side of the equation is the perfect trinomial square which can be framed as ${(x - a)^2}$
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \left( {\dfrac{1}{4}} \right)$
Take the square root on both the sides of the equation.
$ \Rightarrow \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2}} = \sqrt {\left( {\dfrac{1}{4}} \right)} $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow \left( {x - \dfrac{3}{2}} \right) = \pm \sqrt {\left( {\dfrac{1}{4}} \right)} $
Make the required term as the subject –
$ \Rightarrow x = \dfrac{3}{2} \pm \sqrt {\dfrac{1}{4}} $
Simplify the above expression –
$
   \Rightarrow x = \dfrac{3}{2} \pm \dfrac{1}{2} \\
   \Rightarrow x = \dfrac{3}{2} + \dfrac{1}{2}{\text{ or }}x = \dfrac{3}{2} - \dfrac{1}{2}{\text{ }} \\
   \Rightarrow x = 2\,{\text{or x = 1}} \\
 $
This is the required solution.

Note: Always remember that the square root of positive term can be positive or negative term while the square of positive or negative term always gives you the positive term. Be good in multiples and squares and square root concepts.