Question

How do you solve the quadratic equation by completing the square: \[{{x}^{2}}-6x=0\]

Answer 1

Hint: This question belongs to the topic quadratic equation. In this question, first we will add the square of half coefficient of x to the both sides of the equation. After that, we will make the equation in the form of \[{{a}^{2}}-2ab+{{b}^{2}}\]. As we know that the equation \[{{a}^{2}}-2ab+{{b}^{2}}\] is the perfect square of \[\left( a-b \right)\]. So, we will use this formula and solve the further solution to get the value of x.

Complete step by step answer:
Let us solve this question.
In this question, we have to solve the quadratic equation \[{{x}^{2}}-6x=0\] by completing the square.
For solving this question by completing the square, firstly we will make sure that the coefficient of \[{{x}^{2}}\] is 1. We can see in the equation \[{{x}^{2}}-6x=0\] that coefficient of \[{{x}^{2}}\] is 1. Now, we can solve further.
In the given equation, we are going to add the square of half of coefficient of x (that is 9) to the both of equation. We will get,
\[{{x}^{2}}-6x+9=0+9\]
The above equation can also be written as
\[{{x}^{2}}-2\times 3\times x+{{3}^{2}}={{3}^{2}}\]
As we can see that left hand side of equation is in the form of \[{{a}^{2}}-2ab+{{b}^{2}}\] and we know that the equation \[{{a}^{2}}-2ab+{{b}^{2}}\] is equal to \[{{\left( a-b \right)}^{2}}\]. By comparing both the equations, we can say that a=x and b=3. So, we can write the above equation as
\[\Rightarrow {{\left( x-3 \right)}^{2}}={{3}^{2}}\]
Now, taking the square root to both the side of equation, we get
\[\Rightarrow \sqrt{{{\left( x-3 \right)}^{2}}}=\pm \sqrt{{{3}^{2}}}\]
We can write the above equation as
\[\Rightarrow \left( x-3 \right)=\pm 3\]
We can write the equation as
 \[\Rightarrow x=3\pm 3\]
From the above equation, we have got the two values of x.
x=3-3 and x=3+3

From the above, we can say that the values of x are 0 and 6.

Note: We should have a better knowledge in the topic quadratic equation. Don’t forget to give both plus and minus signs while taking the square root of a number. Remember the following formula to solve this type of question easily:
\[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]
 We have a different method to solve this question.
As we have found from the above that
\[{{\left( x-3 \right)}^{2}}={{3}^{2}}\]
We can solve from here by a different method.
The above equation can also be written as
\[\Rightarrow {{\left( x-3 \right)}^{2}}-{{3}^{2}}=0\]
As we know that \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]. Using this formula in the above by putting a=x-3 and b=3, we can write
\[\Rightarrow \left( x-3-3 \right)\left( x-3+3 \right)=0\]
The above equation can also be written as
\[\Rightarrow \left( x-6 \right)\left( x+0 \right)=0\]
From here, we can say that
x=-6 and x=0
We get the same solution. So, we can use this method too.