Question

How do you solve ${x^2} - 3x + 6 = 0$ by completing the square?

Answer 1

Hint: To solve the given equation using completing the square method, we start with the middle term of the equation, evaluate it in a way to convert it into Perfect Square Trinomial. Then simplify further until you get the desired result.

Complete Step by Step Solution:
Firstly, do this only when the numerical coefficient of ${x^2}$ is 1.
Secondly, start with the numerical coefficient of x which is the number -3.
Then divide this number by 2 and then square the result.
That is: ${\left( {\dfrac{{ - 3}}{2}} \right)^2} = \dfrac{9}{4}$
Add $\left( {\dfrac{9}{4}} \right)$ to both the sides of the equation we will get:
$ \Rightarrow {x^2} - 3x + \dfrac{9}{4} + 6 = 0 + \dfrac{9}{4}$
The first three terms now become one group which is a PST- Perfect Square Trinomial.
Simplify by putting brackets to the right place we get:
$ \Rightarrow \left( {{x^2} - 3x + \dfrac{9}{4}} \right) + 6 = 0 + \dfrac{9}{4}$
Now, after using the formula ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$in the above equation we will see:
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} + 6 = \dfrac{9}{4}$
After transposing the 6 to the right hand side:
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{9}{4} - 6$
Now, we will take the LCM of the denominators in the right hand side we get 4:
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{9 - 24}}{4}$
Taking the under root on both the side of the equation:
$ \Rightarrow \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2}} = \pm \sqrt {\dfrac{{9 - 24}}{4}} $
Simplify:
$ \Rightarrow x - \dfrac{3}{2} = \pm \sqrt {\dfrac{{ - 15}}{4}} $
Separating the under root on numerator and denominator:
$ \Rightarrow x - \dfrac{3}{2} = \pm \dfrac{{\sqrt { - 15} }}{{\sqrt 4 }}$
We know that the under root of 4 is 2 so will insert the value in the above equation
$ \Rightarrow x - \dfrac{3}{2} = \pm \dfrac{{\sqrt { - 15} }}{2}$
Finally transpose $ - \dfrac{3}{2}$to the right side of the equation
$ \Rightarrow x = \dfrac{3}{2} \pm \dfrac{{\sqrt { - 15} }}{2}$
Take note here: $\sqrt { - 15} = \sqrt {15} \cdot \sqrt { - 1} = \sqrt {15} i$
After substituting the value of $\sqrt { - 15} $we will get:
$ \Rightarrow x = \dfrac{3}{2} \pm \dfrac{{\sqrt {15} i}}{2}$

Therefore, there are two values of x:$ \Rightarrow x = \dfrac{{3 + \sqrt {15} i}}{2},\dfrac{{3 - \sqrt {15} i}}{2}$

Note: Always use completing the square method when the numerical coefficient of ${x^2}$is 1.
2. Remember to use the formula ${\left( {\dfrac{b}{2}} \right)^2}$ in order to create a new term.
3. solve for x by using this term to complete the square.