
Paul H. answered 09/20/13
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Math, Excel, InDesign, Photography and Computers in General
Hi Harry,
This equation can be solved by factoring.
When i am doing these, i like to think of all the possible factors of the coefficient in front of x^2.....in this case 24.
Here is what they could be:
1 x 24
2 x 12
3 x 8
4 x 6
then, i do the same for the constant (3 in this case). Since 3 is a prime number, there is only one set of factors:
1 x 3
now the "hard" part is choosing which factors to use to get the middle coefficient (-14 in this case)
Since it is negative, we know that we will need a + and a - in the two factors:
(?x + ?)(?x - ?)
a little experimenting and we can get:
(6x + 1)(4x - 3) = 0
If we generalize this equation to be A * B = 0. So if we make A = 0 then B doesn't matter because 0 * anything = 0. Same idea for B. So what we do next is to make each factor a little mini equation like this:
(6x + 1) = 0
6x = -1
x = -1/6
(4x-3) = 0
4x = 3
x = 3/4
so the roots are:
x=3/4 or x = -1/6
Does that make sense?
Now, the harder part.....can every quadratic equation be factored? I believe the short answer is yes, but it may be very difficult if the coefficients are not easy numbers. However, there are different methods that you could use to figure it out (like completing the square and the quadratic formula ). These methods are not as simple, but will work regardless of how "easy" the factors are.