Completing the square is not the easiest way to solve quadratic equations; its strength lies in the fact that the process is repetitive and predictable.
Here is the best news: completing the square ALWAYS (SAY ALWAYS) will work, unlike the factoring method, which of course, requires that the trinomial be factorable.
What are the pros and cons of completing the square as a way to solve quadradic equations?
6 answers
I find the quadratic equation
x = [-b +/- sqrt(b^2-4ac)]/2a
the easiest to use, unless a way of factoring is obvious. It is derived by completing the square, after all. It tells you the number of real roots right away (from the value of b^2 - 4ac). The hard part is memorizing it, but after a while it becomes routine.
x = [-b +/- sqrt(b^2-4ac)]/2a
the easiest to use, unless a way of factoring is obvious. It is derived by completing the square, after all. It tells you the number of real roots right away (from the value of b^2 - 4ac). The hard part is memorizing it, but after a while it becomes routine.
I use the following rule:
If the coefficient of the squared term is 1 and the coefficient of the first degree term is even, then I would use completing the square, otherwise just use the quadratic formula
e.g.
x^2 - 12x -5 = 0
x^2 - 12x = 5
x^2 - 12x + 36 = -5 + 36
(x-6)^2 = 31
x = 6 ± √31
In this case this method is actually faster and easier than using the formula, since the formula answer has to be broken down to lowest terms.
If the coefficient of the squared term is 1 and the coefficient of the first degree term is even, then I would use completing the square, otherwise just use the quadratic formula
e.g.
x^2 - 12x -5 = 0
x^2 - 12x = 5
x^2 - 12x + 36 = -5 + 36
(x-6)^2 = 31
x = 6 ± √31
In this case this method is actually faster and easier than using the formula, since the formula answer has to be broken down to lowest terms.
Thank you.
Cool rule, thank you.
Thank you for the example.
1 answer