
How do you solve \[{{a}^{2}}-a-1=0\] by completing the square?
Answer
548.4k+ views
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, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
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, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

A moving boat is observed from the top of a 150 m high class 10 maths CBSE

