Question

How do you solve \[{{x}^{2}}-5x-10=0\] by completing the square?

Answer 1

Hint: For solving this question, first we will add the square of the half coefficient of x term to both sides of the equation. After that, we will arrange the equation. After that, we will see that one side of equation will be of the form \[{{a}^{2}}+2ab+{{b}^{2}}\] or \[{{a}^{2}}-2ab+{{b}^{2}}\] . As we know that \[{{a}^{2}}+2ab+{{b}^{2}}\] and \[{{a}^{2}}-2ab+{{b}^{2}}\] is a perfect square which is equal to \[{{\left( a+b \right)}^{2}}\] and \[{{\left( a-b \right)}^{2}}\] respectively. So, we will use these formulas for solving this question. From here, we will find the value of x by solving the further equation.

Complete step by step answer:
Let us solve this question.
This question is asking us to solve \[{{x}^{2}}-5x-10=0\] by completing the square.
For solving this equation by completing the square.
Firstly, we have to make sure that the coefficient of \[{{x}^{2}}\] term is 1. As we can see that the coefficient of \[{{x}^{2}}\] is 1 in the given equation.
Now, we will add the square of half of coefficient of x term to both sides of the equation \[{{x}^{2}}-5x-10=0\] , we get
\[{{x}^{2}}-5x-10+{{\left( \dfrac{1}{2}\times \left( -5 \right) \right)}^{2}}=0+{{\left( \dfrac{1}{2}\times \left( -5 \right) \right)}^{2}}\]
\[\Rightarrow {{x}^{2}}-5x-10+\left( \dfrac{25}{4} \right)=\left( \dfrac{25}{4} \right)\]
Which is also can be written as
\[\Rightarrow {{x}^{2}}-5x+\dfrac{25}{4}=\dfrac{25}{4}+10\]
\[\Rightarrow {{x}^{2}}-5x+\dfrac{25}{4}=\dfrac{65}{4}\]
Which is also can be written as
\[\Rightarrow {{x}^{2}}-2\times \left( \dfrac{5}{2} \right)\times x+\dfrac{25}{4}=\dfrac{65}{4}\]
As \[\dfrac{25}{4}={{\left( \dfrac{5}{2} \right)}^{2}}\]
Hence
\[\Rightarrow {{x}^{2}}-2\times x\times \left( \dfrac{5}{2} \right)+{{\left( \dfrac{5}{2} \right)}^{2}}=\dfrac{65}{4}\]
Now, we can see that the left side of equation making the perfect square in the form of \[{{a}^{2}}-2ab+{{b}^{2}}\]\[{{a}^{2}}-2ab+{{b}^{2}}\].
As we know that \[{{a}^{2}}-2ab+{{b}^{2}}\] is also can be written as \[{{\left( a-b \right)}^{2}}\].
Therefore, \[{{x}^{2}}-2\times x\times \left( \dfrac{5}{2} \right)+{{\left( \dfrac{5}{2} \right)}^{2}}\] can be written as \[{{\left( x-\dfrac{5}{2} \right)}^{2}}\]
So, the equation \[{{x}^{2}}-2\times x\times \left( \dfrac{5}{2} \right)+{{\left( \dfrac{5}{2} \right)}^{2}}=\dfrac{65}{4}\] can be written as
\[\Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{65}{4}\]
Now, taking the square root of both sides, we get
\[\Rightarrow \sqrt{{{\left( x-\dfrac{5}{2} \right)}^{2}}}=\pm \sqrt{\dfrac{65}{4}}\]
\[\Rightarrow \left( x-\dfrac{5}{2} \right)=\pm \sqrt{\dfrac{65}{4}}=\pm \dfrac{\sqrt{65}}{2}\]
\[\Rightarrow \left( x-\dfrac{5}{2} \right)=\pm \dfrac{\sqrt{65}}{2}\]
Now, adding both the side \[\dfrac{5}{2}\], we get
\[\Rightarrow x=\dfrac{5}{2}\pm \dfrac{\sqrt{65}}{2}\]
Hence, we get the value of x by completing the square.

Note:
 For solving this type of question, we should have a better knowledge in polynomials and in quadratic equations. Also, remember the formulas like:
\[\begin{align}
  & {{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}} \\
 & {{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}} \\
\end{align}\]
And, don’t forget to put the signs of plus (+) and minus (-) whenever finding the root of a constant.
We can see an example taken from the above solution.
\[\Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{65}{4}\]
Now, taking the square root of both sides, we get
\[\Rightarrow \sqrt{{{\left( x-\dfrac{5}{2} \right)}^{2}}}=\pm \sqrt{\dfrac{65}{4}}\]
Here, if we write \[\sqrt{\dfrac{65}{4}}\] in the place of\[\pm \sqrt{\dfrac{65}{4}}\], then the solution will be wrong. So, one should get informed of that.