Lesson HOW TO solve quadratic equation by completing the square

This Lesson (HOW TO solve quadratic equation by completing the square - Learning by examples) was created by by ikleyn(52781)

: View Source, Show
About ikleyn:

HOW TO solve quadratic equation by completing the square - Learning by examples

It was just described in the lesson PROOF of quadratic formula by completing the square how to solve quadratic equation by completing the square.
That explanation was theoretical and included many expressions and expression transformations.

Some students prefer to learn by examples.
Explanations in this lesson are designed for such students.

Problem 1

Solve for x by completing the square x%5E2+%2B+8x+-4

.

Below two methods (two solutions) are presented in two columns, and you will clearly see that they are equivalent.

Solution 1

x^2 + 8x - 4 = 0.

Look in these two terms containing x:  . 
So, the first thought which comes to your mind is to get (x+4)^2.
To bring it to reality, you need to add  =  to the left side      
polynomial and to distract it to keep that polynomial expression 
unchanged. Now, see how I do it below:


     = 

    is equivalent to

     = 

    is equivalent to

     - 20 = 

    is equivalent to

     = 20

    is equivalent to

    x+4 = +/- 

    is equivalent to

    x = .

Solution 2

x^2 + 8x - 4 = 0.

Group these two terms containing x,  , in the left side. 
Move the constant term  to the right side with the opposite sign.
Next, transform the left side to .
For it, add  =  to both sides. Now, see how I do it below:


     = 

    is equivalent to

     = 

    is equivalent to

     = 

    is equivalent to

     = 

    is equivalent to

    x+4 = +/- 

    is equivalent to

    x = .

The two roots are x%5B1%5D

and

.
The solution procedure by the method of completing the square is done.

The two versions of the procedure are presented, and they are equivalent by the obvious way.
You may use either of the two versions.

Problem 2

Solve for x by completing the square x%5E2-10x%2B3

.

Solution

 = 0

is equivalent to

x^2 - 10x = -3.

Take half of the coefficient (-10) at "x" and add the term  to both sides of the equation. You will get an equivalent equation
 = ,   or

 = .

The left side is . Therefore, the last equation takes the form
 = 22.

Take square root of both sides. You will get

x-5 = +/- ,   or, which is the same,

x = .

The two roots are  =   and   = .

The solution is completed.  The procedure is the same as in the solution to the Problem 1.

Problem 3

Solve an equation by completing the square 2x%5E2%2B5x%2B3

.

Solution

  = .

Divide both sides of the equation by 2:

 = .

Subtract  from both sides of the equation:

 = 

Divide  by 2 and square it, , add  to both sides of the equation:

 =  = .

 = .

Take square root of both sides of the equation:

 = +/- .

Now we have two solutions for x

    x =  = -1     and   x =  = -1.5.


The solution is completed.  The procedure is the same as in the solution to the Problem 1  and Problem 2.

Problem 4

Solve by completing the square -3x%5E2%2B1

.

Solution

 =   <---->   =   <--->   

 =   <---->   =   <---->   =  <---->  

 =   <---->   =   <---->   =   <---->

 = +/-   <---->  x =   ---->


There are two solutions:   =  = 0.215 (approx.),   and    =  = -1.55 (approx.).