How do you solve \[{{a}^{2}}-a-1=0\] by completing the square?

Hint:In the given question, we have been asked to find the value of ‘a’ by solving the given equation i.e. \[{{a}^{2}}-a-1=0\] using the completing the square method. Completing the square method is used to solve the quadratic equation by converting the form of the equation so that it will become a perfect trinomial.

Complete step by step solution:
We have given that,
\[\Rightarrow {{a}^{2}}-a-1=0\]
Adding 1 to both the sides of the equation, we get
\[\Rightarrow {{a}^{2}}-a-1+1=0+1\]
Simplifying the above equation, we get
\[\Rightarrow {{a}^{2}}-a=1\]
Now, we completing the square, we add \[{{\left( \dfrac{1}{2} \right)}^{2}}\]to both the sides of the equation, we get
\[\Rightarrow {{a}^{2}}-a+{{\left( \dfrac{1}{2} \right)}^{2}}=1+{{\left( \dfrac{1}{2} \right)}^{2}}\]
Simplifying the above equation, we get
\[\Rightarrow {{a}^{2}}-a+{{\left( \dfrac{1}{2} \right)}^{2}}=1+\left( \dfrac{1}{4} \right)\]
\[\Rightarrow {{a}^{2}}-a+{{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{5}{4}\]
As we know that, \[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]
Therefore, applying this identity we get
\[\Rightarrow {{\left( a-\dfrac{1}{2} \right)}^{2}}=\dfrac{5}{4}\]
Transposing the power 2 on the right side of the equation, we get
\[\Rightarrow \left( a-\dfrac{1}{2} \right)=\sqrt{\dfrac{5}{4}}\]
As we know that any root value takes two values i.e. one positive value and the other is negative value.
Simplifying the above equation, we get
\[\Rightarrow \left( a-\dfrac{1}{2} \right)=\pm \dfrac{\sqrt{5}}{2}\]
Adding \[\dfrac{1}{2}\] to both the sides of the equation, we get
\[\Rightarrow a=\pm \dfrac{\sqrt{5}}{2}+\dfrac{1}{2}\]
Now, we have
\[\Rightarrow a=\dfrac{1+\sqrt{5}}{2}\ or\ \dfrac{1-\sqrt{5}}{2}\]
Therefore, the possible values of ‘a’ are \[\dfrac{1+\sqrt{5}}{2}\ and\ \dfrac{1-\sqrt{5}}{2}\].
Formula used:
\[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]

Note: While solving these types of questions, students should carefully observe when considering the terms, which one is the ‘a’ and the ‘b’. To check whether the obtained possible values are correct or not, we can verify the result by solving the quadratic equation with the roots of the quadratic equation formula. Standard form of quadratic equation; \[a{{x}^{2}}+bx+c=0\], then the roots of the quadratic equation is given by, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].