Question

How do you solve ${x^2} + 6x - 4 = 0$ by completing the square?

Answer 1

Hint: Here, we will add and subtract the square of a constant and write the term containing $x$ such that it forms a term in the form $2ab$. We will then apply the square identities and ‘complete the square’ and solve it further to find the required roots of the given quadratic equation.

Formula Used:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.

Complete step-by-step answer:
The given quadratic equation is ${x^2} + 6x - 4 = 0$.
Now, since, we are required to solve this question using completing the square, hence, we will write this quadratic equation as:
${\left( x \right)^2} + 2\left( x \right)\left( 3 \right) + {\left( 3 \right)^2} - {\left( 3 \right)^2} - 4 = 0$.
Hence, even if this quadratic equation was not a perfect square, but we tried to make it a perfect square by adding and subtracting the square of a constant such that, we complete the square by making the identity, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
Using this identity, we get,
$ \Rightarrow {\left( {x + 3} \right)^2} - 9 - 4 = 0$
Adding the like terms, we get
$ \Rightarrow {\left( {x + 3} \right)^2} - 13 = 0$
Adding 13 on both the sides, we get
$ \Rightarrow {\left( {x + 3} \right)^2} = 13$
Thus, taking square root on both sides of ${\left( {x + 3} \right)^2} = 13$, we get,
$\left( {x + 3} \right) = \pm \sqrt {13} $
Subtracting 3 from both the sides, we get
$ \Rightarrow x = - 3 \pm \sqrt {13} $
Thus, this is the required answer.

Note: If in the question, it was not mentioned that we have to use the method of completing the square, then, we could have used the quadratic formula to solve the given quadratic equation.
Given quadratic equation is ${x^2} + 6x - 4 = 0$
Comparing this with the general quadratic equation i.e. $a{x^2} + bx + c = 0$.
We have,
$a = 1$, $b = 6$and $c = - 4$
Now, we can find the roots of a quadratic equation using the quadratic formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Hence, for the equation ${x^2} + 6x - 4 = 0$, substituting \[a=\] , $b = 6$ and $c = - 4$, we get,
$x = \dfrac{{ - 6 \pm \sqrt {{6^2} - 4\left( 1 \right)\left( { - 4} \right)} }}{{2\left( 1 \right)}}$
Simplifying the expression, we get
$ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 + 16} }}{2} = \dfrac{{ - 6 \pm \sqrt {52} }}{2}$
Solving further, we get,
$ \Rightarrow x = \dfrac{{ - 6 \pm 2\sqrt {13} }}{2}$
Dividing numerator and denominator by 2, we get
$ \Rightarrow x = - 3 \pm \sqrt {13} $
Thus, this is the required answer.