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Completing the square is a standard topic in algebra II, and is assumed for this course. So you might need to review the process, which I attempt to explain below. At the end of this note I also give you a reference in your text, where you'll see an alternative explanation.
y^2 + 12 y is not the square of any binomial.
y^2 + 12 y + 36 is the square of the binomial (y + 6), as you should verify.
Also y^2 + 12 y + 36 is the only quadratic trinomial containing the terms y^2 + 12 y which is the square of a binomial.
Of course y^2 + 12 y + 36 isn't the same as y^2 + 12 y, but y^2 + 12 y + 36 - 36 is the same, since 36 - 36 = 0.
So we can write
y^2 + 12 y as
y^2 + 12 y + 36 - 36,
and it doesn't change anything if we group this expression as
(y^2 + 12 y + 36) - 36.
Hopefully you've verified that y^2 + 12 y + 36 is the same as (y + 6)^2, as was suggested above.
So
(y^2 + 12 y + 36) - 36
is the same as
(y + 6)^2 - 36.
So we conclude that
y^2 + 12 y = (y + 6)^2 - 36.
Now the question is how we could have figured that out without already knowing it.
Remember that in general
(a + b)^2 = a^2 + 2 a b + b^2,
as you should verify.
To get the idea, use the formula to find the following:
(y + 1)^2
(y + 2)^2
(y + 3)^2.
(y + 4)^2.
OK, you should have found all four of these, but let's think about (y + 4)^2. As you should now know
(y + 4)^2 = y^2 + 8 y + 16.
The number in front of y is 8 and the constant term in the expression is 16. What do these numbers have in common?
The answer is that they both come from the 4 in (y + 4)^2. The 8 is 2 times 4 (related to the term 2 a b in the formula), and the 16 is 4^2 (related to the term b^2 in the formula).
Now if we were to see
y^2 + 8 y
we could look at it like this. y^2 + 8 y is the beginning of a perfect square, the square of y plus some 'mystery number'. 8 y is the 2 a b term of our square, and a stands for y. So 2 b must stand for 8, and b must therefore be 4.
Our conclusion:
y^2 + 8 y is the beginning of the square (y + 4)^2. Check this out.
Sure enough,
(y + 4)^2 = y^2 + 8 y + 16.
In our problem we started with y^2 + 12 y. The 12 is the 2 b of our 2 a b term, so b = 6.
Thus y^2 + 12 y is the beginning of the square of (y + 6). To verify once more,
(y + 6)^2 = y^2 + 12 y + 36.
Hopefully you understand this. Whether you do or not, you should go to your text and review 'completing the square'. Look it up in the index; you will see references to the Appendix, in the review section. In my text it's on pages A29-30 and A48-49.
A number of examples in your text use the process of completing the square.
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As to why we complete the square, we do so to get the equation into the standard form
x - h = 1 / (4 a) * (y - k)^2.
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