Question

How do you solve the equation ${x^2} + 2x - 6 = 0$ by completing the square ?

Answer 1

Hint: This problem deals with solving the given quadratic equation with the completing the square method. Completing the square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial. To solve $a{x^2} + bx + c = 0$ by completing the square: transform the equation so that the constant term,$c$ is alone on the right side.

Complete step-by-step solution:
Given a quadratic equation and we are asked to solve the quadratic equation in the completing the square method.
The given quadratic equation is given by: ${x^2} + 2x - 6 = 0$
Now consider the given equation as shown below:
$ \Rightarrow {x^2} + 2x - 6 = 0$
Add 1 on both sides to the equation as shown below:
$ \Rightarrow {x^2} + 2x + 1 - 6 = 1$
$ \Rightarrow \left( {{x^2} + 2x + 1} \right) - 6 = 1$
Now move the number -6 on to the other side of the equation, as ${x^2} + 2x + 1$, makes a perfect square:
$ \Rightarrow {x^2} + 2x + 1 = 7$
$ \Rightarrow {\left( {x + 1} \right)^2} = 7$
Now take square root on both sides as shown:
$ \Rightarrow \left( {x + 1} \right) = \pm \sqrt 7 $
So the solutions of $x$ are:
$ \Rightarrow \left( {x + 1} \right) = \sqrt 7 $ and $\left( {x + 1} \right) = - \sqrt 7 $
Hence $x = - 1 + \sqrt 7 $ and $x = - 1 - \sqrt 7 $
So the solutions of are as given below:
$\therefore x = - 1 \pm \sqrt 7 $

The solutions of $x$ are $ - 1 \pm \sqrt 7 $.

Note: While solving this problem please be careful when considering the terms, which of the one will be $a$, and which one will be $b$. To check whether the obtained result is correct or not, we can verify the result by solving the quadratic equation with the roots of the quadratic equation formula. If the quadratic equation is represented by : $a{x^2} + bx + c = 0$, then the roots of the quadratic equation is given by the formula : $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.