Question

Question #4fa4f

Answer 1

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+6x+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+6x=-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+6x+9=-1+9`

`=> x^2+6x+9=8`

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

`(x+3)^2=8`

Take square root of both sides and remember to use both `+ and -` signs on one side. We did it for RHS.

`sqrt((x+3)^2)=sqrt8`
`(x+3)=+-sqrt8`

Solve for `x`
`x=-3+-sqrt8`
or `x=-3+-2sqrt2`

The two roots of the given equation are
`x=-3+2sqrt2` and `x=-3-2sqrt2`