Biology Reference
In-Depth Information
224
QUADRATIC ALGEBRAIC EQUATION
Quadratic equations appear frequently in many mathematical problems. The
two solutions can be obtained using some simple algebraic relations.
Consider:
A x
2
+ B x + C = 0,
where A, B and C are constants. Dividing the equation by A gives:
x
2
+ (B/A) x + (C/A) = 0 .
Then, add on a term so that the first two terms can be combined to form a
perfect square, i.e.
x
2
+ (B/A) x + (B/2A)
2
- (B/2A)
2
+ (C/A) = 0,
{x + (B/2A)}
2
- (B
2
- 4AC)/(2A)
2
= 0,
or
or {x + (B/2A) - (B
2
- 4AC)
l/2
/2A} {x + (B/2A) + (B
2
- 4AC)
1/2
/2A} = 0 .
Therefore, we get:
x = {-B + (B
2
- 4AC)
1/2
}/2A,
x = {-B - (B
2
- 4AC)
1/2
}/2A .
or
This result is used to solve for
in the Pauling's model of hemoglobin
saturation (see Chapter 2).
