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Question: How do you solve \({x^2} - 3x + 6 = 0\) by completing the square?...

How do you solve x2−3x+6=0{x^2} - 3x + 6 = 0 by completing the square?

Explanation

Solution

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 x2{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: (−32)2=94{\left( {\dfrac{{ - 3}}{2}} \right)^2} = \dfrac{9}{4}
Add (94)\left( {\dfrac{9}{4}} \right) to both the sides of the equation we will get:
⇒x2−3x+94+6=0+94\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:
⇒(x2−3x+94)+6=0+94\Rightarrow \left( {{x^2} - 3x + \dfrac{9}{4}} \right) + 6 = 0 + \dfrac{9}{4}
Now, after using the formula a2−2ab+b2=(a−b)2{a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}in the above equation we will see:
⇒(x−32)2+6=94\Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} + 6 = \dfrac{9}{4}
After transposing the 6 to the right hand side:
⇒(x−32)2=94−6\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:
⇒(x−32)2=9−244\Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{9 - 24}}{4}
Taking the under root on both the side of the equation:
⇒(x−32)2=±9−244\Rightarrow \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2}} = \pm \sqrt {\dfrac{{9 - 24}}{4}}
Simplify:
⇒x−32=±−154\Rightarrow x - \dfrac{3}{2} = \pm \sqrt {\dfrac{{ - 15}}{4}}
Separating the under root on numerator and denominator:
⇒x−32=±−154\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
⇒x−32=±−152\Rightarrow x - \dfrac{3}{2} = \pm \dfrac{{\sqrt { - 15} }}{2}
Finally transpose −32 - \dfrac{3}{2}to the right side of the equation
⇒x=32±−152\Rightarrow x = \dfrac{3}{2} \pm \dfrac{{\sqrt { - 15} }}{2}
Take note here: −15=15⋅−1=15i\sqrt { - 15} = \sqrt {15} \cdot \sqrt { - 1} = \sqrt {15} i
After substituting the value of −15\sqrt { - 15} we will get:
⇒x=32±15i2\Rightarrow x = \dfrac{3}{2} \pm \dfrac{{\sqrt {15} i}}{2}

Therefore, there are two values of x:⇒x=3+15i2,3−15i2 \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 x2{x^2}is 1.
2. Remember to use the formula (b2)2{\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.