Solving a quadratic by completing the square when we have a coefficient on our leading term. Whenever we have a coefficient on our leading term what we’re always going to have to do is factor out after we isolate our x variables. The first thing we want to do is subtract that 2 over and then after we do that we want to get the leading coefficient on our xs, our x² term to be 1. So we need to factor out that 2, end up with 2x minus 3x is equal to -2.
Now we want to somehow express these parentheses as a perfect square. We know we have an x it’s going to have to be a minus. We’re trying to figure out what would give us a 3 in the middle term. The way we always do that is this term divided by 2 is what goes here and then that squared is what goes up over here, so this ends up being plus 9/4.
We still have the 2 coefficient on the outside, and in order to keep it balanced we have to figure what actually added to this side as a whole. We added 9/4 inside the parentheses but we have this 2 on the outside okay, so we need to make sure that when we add that to the other side we add the exact same thing that we added. 2 times 9/4 is 9/2 so we really added 9/2 to this side as a whole, keeping it balanced, I have to add 9/2 over here as well.
We have 2, x minus 3/2 squared is equal to -2 is -4 halves so this ends up being 5/2. We want to isolate the square so we need to divide by 2 which is the same thing as multiplying by ½ leaving us with x minus 3/2 squared is equal to 5/4.
Finishing this up we need to take the square root of both sides, remember whenever we take this square root as a tool we have to include plus or minus so what we end up with is x minus 3/2 is equal to plus or minus root 5/4. Add 3/2 to both sides and in the same step I’m going to simplify this square root of a fraction is the square root of the top over the square root of the bottom. Square root of 4 is just 2. What we end up with is x is equal to 3/2 plus or minus root 5 over 2. If you wanted to take this and write it over a single fraction you could, we already have a common denominator but no need, in most cases.
Really what we did is solve the quadratic by completing the square, we had a coefficient, our middle term it was an odd number which means we’re left with a fraction but the same process holds, numbers a little bit uglier but we went through the whole entire process. And the last thing I want to point out is always make sure that when you have that coefficient you distribute that term through okay, to make sure you keep each side balanced when you add that in.