How do you use the quadratic complement?
Hello,
I understand the quadratic completion from the beginning (i.e. factoring out the a) but then the further steps not at all!
I've watched countless explanatory videos, textbook explanations, notebook entries, and answers to Gutefrage, but I still don't understand any of them. It's complicated everywhere, and I can't follow the steps.
Maybe I can find someone here who can explain it to me step by step in a less complicated way.
thanks in advance
I hope you still know the binary formula (a + b)2 = a2 + 2*a*b + b2 (*).
The aim of the square supplement is to create a square polynomial term
g(x) = a*x2 + b*x + c
into a binomus and into a rest that does not contain x. For the sake of simplicity (Mathematicians call this “without restriction of the generality” or, above all, of which a has already been excluded, so we have a term
h(x) = x2 + p*x + q (**)
have. In order to add the subterm x2 + p*x to a binomin (therefore the term “addition”), it must first be made from the p a 2*(p/2):
x2 + p*x = x2 + 2*(p/2)
Now (p/2)2 is “added” and we receive
x2 + 2*(p/2)*x + (p/2)2 = (x + p/2)2 (***)
So we have found the value with the “added”. Now comes a trick that will always be important. We re-form the term in (**) by using a +0 and we do so by adding the subterm p/2 – p/2) (= 0) to (**). Consider yourself why this does not change the value of (**):
h(x) = x2 + p*x + q = x2 + p*x + p/2 – p/2 + q = x2 + 2*(p/2) + p/2 – p/2 + q
Now we can use the relationship found in (***) and finally get
h(x) = x2 + 2*(p/2) + p/2 – p/2 + q = (x + p/2)2 – p/2 + q
We have thus achieved the objective set out above. We have a binome (x+p/2)2 and a subterm that no longer contains x – p/2 + q
Why is that useful? The equation x2+p*x+q=0 can first be solved with simple deformations. These deformations lead to the so-called pq formula (and if a a is present to the so-called abc or midnight formula). Furthermore, the so-called apex of the parabola h(x) can be easily read from the equation of the square addition, namely, it lies with
xs = -p/2, ys = -p/2 + q
xs is precisely the value at which the term within the bracket becomes 0, ys is the lowest or highest point of the parabola (depending on whether a > 0 or a < 0).
First try to understand the calculation steps. But then I can only give you the recommendation to learn the procedure stubbornly and not to draw you back to a “why must I do it”. Because that only leads to confusion and frustration.
Output form x2 + b*x + c
1. Step
x2 + b*x + c + (b/2)2 – (b/2)2
Two. Step
[ x2 + b*x + (b/2)2 ] + c – (b/2)2
Mould the expression in the square bracket:
(x + b/2)2 + c – (b/2)2
3. Step
(x2 + b/2)2 + d, with d = c – (b/2)2
finished
Sounds the initial form a*x2 + b*x + c, with a != 1
you divide all terms by a
x2 + b/a*x + c/a
Then you keep going up.
In the end all terms are multiplied by a:
a*(x2 + b/2)2 + a*d
###
Example:
1/2*x2 – 4x + 5
Factor a=1/2 eliminate:
x2 – 8x + 10
1. Step (b/2 = -8/2 = -4)
x2 – 8x + 10 + (-4)2 – (-4)2
Two. Step
[x2 – 8x + (-4)2] + 10 – (-4)2
is identical to
(x – 4)2 + 10 – (-4)2
3. Step
(x – 4)2 + 10 – 16 = (x – 4)2 – 6
Now again take into account the factor a = 1/2:
1/2*(x – 4)2 – 1/2*6 = 1/2*(x – 4)2 – 3