Mykola V. answered 12/20/12
When we're asked to solve by factoring, we think of quadratic equations. We have all the necessary components here to make one, they're just scattered. Let's first bring all the terms to one side, preferably to the right so our t2 term remains positive.
0=4t2-6t-9+1 -> 0=4t2-6t-8.
We now need to find all the factors of 8 and 4.
8: 1 and 8, 2 and 4.
4: 1 and 4, 2 and 2.
After this, we can use OI of FOIL to figure out whether or not any combination of these will give us 6 when we add or subtract the O and I terms. i.e (2t 2)(2t 4) = 8t 4t. Any sign we put in between these two will not give us 6t. You can try the rest on your own but clearly this method won't work. So we move on to our back plan, the quadratic formula. A good way to memorize it is to sing it to the "Pop goes the weasel" tune. But more to the point, this is the quadratic formula:
x=(-b±√b2-4ac)/2a
As you may know every quadratic formula of degree 2 will have the form 0=ax2+bx+c. This is where we get our a, b and c values. Now to plug them in and solve for x!
x=(6±√(-6)2-4(4)(-8))/2(4)
x=(6±√36+128)/8
x=(6±√164)/8. We can simplify this slightly by manipulating the √164. First we split it into √4√41 and then solve √4=2. Our result:
x=(6±2√41)/8. We can simplify further by dividing the top and bottom by 2.
x=(3±√41)/4 and that is our final answer!
I hope this helped!
Mykola V.
I find that a lot of the students I've had over the years don't remember how to complete the square or just get confused by it a lot. So I just refer them to this method so as to avoid any confusion for the later on.
12/20/12
Rizul N.
The question never asked to solve by the quadratic. That's why I just did it the longer way. I would much rather prefer quadratic (your way) though for this problem just because it's faster and straightforward.
12/20/12