 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
Question from June, a parent: x^2 - 6x = 16 I found you already had this question but the answer you gave did not help me to explain how to answer the problem for my daughter. I do not understand how to complete the square for this particular question and the example you showed did not help. I would appreciate any further help you could give me. |
Hi June,
In my response to this question when Julie sent it I commented that the key is to see the pattern in the squares
(x + a)2 = x2 + 2ax + a2 and
(x - a)2 = x2 - 2ax + a2
The pattern to see is that, in each case if you know the coefficient of x, + 2a or -2a, take half of it and square the result you get the constant term a2. In the problem you sent, on the left side you have x2 - 6x. If you want to expand this to form a square you need to take the coefficient of x, which is -6, divide by 2 to get -6/2 = -3 and then square this number. The square of -3 is 9 so to make x2 - 6x into a square you need to add 9 since x2 - 6x + 9 = (x - 3)2. This is all preliminary work to writing my solution to the problem. The preliminary work was to determine that I need to add 9. here then is how I start my solution
x2 - 6x = 16
add 9 to each side to get
x2 - 6x + 9 = 16 + 9 or
(x - 3)2 = 25 .
Can you complete the problem now?
You sent us two other problems that you can solve similarly. In the second problem you sent , 2x2 - 3x + 1 = 0, there is an extra step. You need the coefficient of x2 to be 1 so divide both sides of the equation by 2 to get
(2x2 - 3x + 1)/2 = 0/2 or
x2 - 3/2 x + 1/2 = 0
and apply the technique above to this expression.
Penny
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.