00:01
The concept involved in this problem is to solve a quadratic equation using completing the square.
00:09
In completing the square, you wish to have, or you should start with, the quadratic and the linear term on one side equation, and the constant term on the other side.
00:21
So in this equation, x squared plus 3 .3x minus 1 .275 equals 0, we are going to leave the x.
00:31
Square plus 3 .3x on the left -hand side of the equation, skip a space, and add the 1 .2775 to each side.
00:48
Now, the reason for skipping that space is we need to insert a number.
00:53
So to get that number, you take the coefficient of the linear term, in this case of 3 .3, divide that by 2, and then square that quotient 3 .3 divided by 2 is 1 .65.
01:19
So squaring that will get us 2 .7225.
01:30
That will be added in this space that we skipped and also to the right -hand side of the equation.
01:45
All right so that brings us to the equation x squared plus 3 .3x plus 2 .725 is equal to and when we add these two numbers together we will get four.
02:08
Okay so the trinomial that is on the left hand side of this equation is a perfect square tritonel...