# Question #4fa4f

May 14, 2016

It is a method to Solve Quadratic Equations .
Please remember that in actual practice unless you're told that you use completing the square method, you will probably not use this method.

#### Explanation:

Suppose you need to solve the following Quadratic equation by completing the square method (I have selected a problem where coefficient of ${x}^{2}$ term is $1$):
${x}^{2} + 6 x + 1 = 0$ ......(1)

First step is to keep all terms containing $x$ on the LHS and take constant terms (not containing $x$) to RHS of the equation. The equation becomes
${x}^{2} + 6 x = - 1$ .....(2)

Now take half of the coefficient of the $x$-term [middle term in equation (1)],
square it, and add that value to both sides of the new equation.

In the example we have coefficient of the $x$-term $= 6$.

Half of $6 = 3$. Square of $3 = 9$. Therefore adding $9$ to both sides of equation (2) we obtain

${x}^{2} + 6 x + 9 = - 1 + 9$

$\implies {x}^{2} + 6 x + 9 = 8$

On inspection we see that we have made LHS a perfect square which can be written as ${\left(x + 3\right)}^{2}$. Doing so the equation becomes

${\left(x + 3\right)}^{2} = 8$

Take square root of both sides and remember to use both $+ \mathmr{and} -$ signs on one side. We did it for RHS.

$\sqrt{{\left(x + 3\right)}^{2}} = \sqrt{8}$
$\left(x + 3\right) = \pm \sqrt{8}$

Solve for $x$
$x = - 3 \pm \sqrt{8}$
or $x = - 3 \pm 2 \sqrt{2}$

The two roots of the given equation are
$x = - 3 + 2 \sqrt{2}$ and $x = - 3 - 2 \sqrt{2}$