Question

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

Answer 1

Hint: Here, we will compare the given quadratic equation with the general quadratic equation to find the coefficient of all the terms. Then we will use the formula of completing the square to add and subtract a constant term such that the given equation is converted into some algebraic identity. Then we will simplify it further to get the required solutions.

Formula Used:
Completing the square method is given by the formula \[{\left( {x \pm \dfrac{b}{2}} \right)^2} + {\left( {\dfrac{b}{2}} \right)^2} = c + {\left( {\dfrac{b}{2}} \right)^2}\]

Complete step by step solution:
We are given with a Quadratic equation \[{x^2} - x - 1 = 0\].
Quadratic equation is of the form \[a{x^2} + bx + c = 0\].
On comparing with the given quadratic equation with the standard form, we get
\[\begin{array}{l}a = 1\\b = - 1\\c = - 1\end{array}\]
Now, we will solve the quadratic equation by completing the square method.
Completing the square method is given by the formula \[{\left( {x \pm \dfrac{b}{2}} \right)^2} + {\left( {\dfrac{b}{2}} \right)^2} = c + {\left( {\dfrac{b}{2}} \right)^2}\]
Now, we will transform the equation in such a way that only the constant term is on the right side of the equation. Therefore, we get
\[{x^2} - x = 1\]
Now, we will add the square of half of the coefficient of \[x\]-term on both the sides of the equation. So, we get
\[ \Rightarrow \left( {{x^2} - x} \right) + {\left( {\dfrac{{ - 1}}{2}} \right)^2} = 1 + {\left( {\dfrac{{ - 1}}{2}} \right)^2}\].
By simplifying the equation, we get
\[ \Rightarrow {x^2} - x + \dfrac{1}{4} = 1 + \dfrac{1}{4}\].
Now, by taking the LCM on the right side of the equation, we get
\[ \Rightarrow {x^2} - x + \dfrac{1}{4} = 1 \times \dfrac{4}{4} + \dfrac{1}{4}\]
\[ \Rightarrow {x^2} - x + \dfrac{1}{4} = \dfrac{4}{4} + \dfrac{1}{4}\]
Adding the terms, we get
\[ \Rightarrow {x^2} - x + \dfrac{1}{4} = \dfrac{{4 + 1}}{4}\]
\[ \Rightarrow {x^2} - x + \dfrac{1}{4} = \dfrac{5}{4}\]
Now using the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\], we get
\[ \Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{5}{4}\]
Now, taking the square root on both sides of the equation, we get
\[ \Rightarrow \sqrt {{{\left( {x - \dfrac{1}{2}} \right)}^2}} = \sqrt {\dfrac{5}{4}} \]
Now, by simplifying the equation, we get
\[ \Rightarrow \left( {x - \dfrac{1}{2}} \right) = \pm \dfrac{{\sqrt 5 }}{2}\]
Now, by rewriting the equation, we get
\[ \Rightarrow x = \dfrac{1}{2} \pm \dfrac{{\sqrt 5 }}{2}\]
\[ \Rightarrow x = \dfrac{1}{2} + \dfrac{{\sqrt 5 }}{2}\] or \[x = \dfrac{1}{2} - \dfrac{{\sqrt 5 }}{2}\]

Therefore, the solution for the Quadratic equation \[{x^2} - x - 1 = 0\] is \[x = \dfrac{1}{2} + \dfrac{{\sqrt 5 }}{2}\] and \[x = \dfrac{1}{2} - \dfrac{{\sqrt 5 }}{2}\].

Note:
We can also solve the quadratic equation by using the method of factorization and the method of Quadratic roots. We should be careful that the quadratic equation should be arranged in the right form. We should also notice that we have both the positive and negative signs in the formula, so the solutions for the equations would be according to the signs. The values of the variable satisfying the given equation are called the roots of a quadratic equation. While performing the completing the square method, we should notice that the coefficient of \[{x^2}\] term \[a\] is 1, if not then divide the entire quadratic equation by \[a\] to get the coefficient of \[{x^2}\] term \[a\] is 1.