We will do a numerical example in parallel with the symbolic form as a concrete illustra-
tion. We start with the general quadratic and factor out thea:
ax^2 +bx+c≡a[
x^2 +b
ax+c
a]
3 x^2 + 5 x− 2 ≡ 3[
x^2 +5
3x−2
3]Now makex^2 +
b
axinto a complete square by adding and subtracting (i.e. adding zero)the square of half the coefficient ofx(b/ 2 a):
(
b
2 a) 2
−(
b
2 a) 2 (
5
6) 2
−(
5
6) 2so that the quadratic becomes
≡a[
x^2 +b
ax+(
b
2 a) 2
−(
b
2 a) 2
+c
a]
3[
x^2 +5
3x+(
5
6) 2
−(
5
6) 2
−2
3]Now note that
x^2 +b
ax+(
b
2 a) 2
=(
x+b
2 a) 2
x^2 +5
3x+(
5
6) 2
=(
x+5
6) 2giving for the original quadratic
≡a[(
x+b
2 a) 2
+c
a−(
b
2 a) 2 ]
3[(
x+5
6) 2
−25
36−24
36]≡a[(
x+b
2 a) 2
+4 ac−b^2
( 2 a)^2]
3[(
x+5
6) 2
−49
36]≡a[(
x+b
2 a) 2
−
( 2 a)^2]where is the discriminant referred to above. /( 2 a)^2 will be either a positive, negative
or zero number, so we can always write the final result in the form
a[(
x+b
2 a) 2
±p^2]
3[(
x+5
6) 2
−(
7
6) 2 ]wherepis some real number.
Note how thea(3 in the example) is retained throughout the calculation, right to the
end. It is a common mistake to drop such factors.
Completing the square in this way gives us the formula for the solution of the quadratic:
ax^2 +bx+c=a[(
x+b
2 a) 2
−b^2 − 4 ac
( 2 a)^2]
= 0