# How do you solve #x^2 = 5x + 2# by completing the square?

##### 3 Answers

#### Answer:

#### Explanation:

First, bring everything to one side...

Complete the square

Solve for

#### Answer:

#### Explanation:

#"rewrite in "color(blue)"standard form"#

#rArrx^2-5x-2=0larrcolor(blue)"in standard form"#

#"to solve by the method of "color(blue)"completing the square"#

#• " the coefficient of the "x^2" term must be 1 which it is"#

#• " add/subtract "(1/2"coefficient of the x-term")^2" to"#

#x^2-5x#

#rArrx^2+2(-5/2)xcolor(red)(+25/4)color(red)(-25/4)-2=0#

#rArr(x-5/2)^2-33/4=0#

#rArr(x-5/2)^2=33/4#

#color(blue)"take the square root of both sides"#

#rArrx-5/2=+-sqrt(33/4)larrcolor(blue)"note plus or minus"#

#"add "5/2" to both sides"#

#rArrx=5/2+-1/2sqrt33larrcolor(red)"exact solutions"#

#### Answer:

Explained the concept of completing the square in a lot of detail.

Once you are used to these you can do them in a LOT LESS lines.

#### Explanation:

Given:

Write as:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We are mathematically allowed to change this in any way we chose as long as we incorporate something that takes it back to the original value. As an example suppose we had:

Let's totally 'lie' about the relationship by writing

This concept is used in completing the square. We lie about something to force it into the format we wish to obtain. After obtaining that format we put back in something that turns it into a true statement.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We need to have

Suppose we make

We can factorise the brackets giving:

If you multiply out the brackets we get

So for this to work we set

Thus

Substitute for

YOU HAVE NOW COMPLETED THE SQUARE

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Set