# Solving quadratic equations: which method to use

```Solving quadratic equations: which method to use?
Once we have been introduced to the quadratic formula, it is tempting to use it to solve any
quadratic equation we come across. While it is true that the quadratic formula will solve ANY
quadratic equation, it is not necessarily the easiest or most efficient method. Refer to page 673
in the text book, and consider the following:
1) Solve 36 x 2  84 x  49  0
This could certainly be solved with the quadratic formula, but the leading coefficient and
the constant should give us a clue that the left side of the equation is, in fact, a trinomial
square. If we see this, we can solve quickly and efficiently without the quadratic
formula:
36 x 2  84 x  49  0
(6 x  7 ) 2  0
6x  7  0
6 x  7
7
x
6
2) Solve x 2  4 x  4  12
Again, the key to doing this problem quickly is to recognize the perfect square trinomial
on the left side. If we recognize it, we can solve the problem more efficiently than
putting it in standard form and using the quadratic equation.
x 2  4 x  4  12
( x  2) 2  12
x  2   12
x  2  12
x  22 3
3) Solve 7 x( x  2)  5  3x( x  1)
With this problem, we don’t know the best approach until we can simplify both sides.
7 x 2  14 x  5  3x 2  3x
4 x 2  11x  5  0
Since neither side results in a “nice” perfect square after the first or second step, this is a
problem best served by using the quadratic formula, with a  4, b  11, c  5 :
 11  112  4(4)(5)  11  121  80  11  41
x


2(4)
8
8
(OVER)
4) Solve 11( x  2)  (5  x)  ( x  2)( x  4)
Notice, this is another one that needs to be simplified on each side first.
11x  22  x  5  x 2  6 x  2 x  12
12 x  27  x 2  4 x  12
0  x 2  16 x  15
At this point we want to see what we have before throwing it into the quadratic formula.
Notice that the right side of the equation will factor very nicely: there is no need to use