|
A quadratic equation is one which may be
expressed in the following general format: Ax2 +
Bx + C = 0 The solution(s)
to the quadratic equation is the following: X = [-B ± √(B2
- 4AC)]/2A x = -B + (B2
– 4AC)1/2 ; x = -B –
(B2 – 4AC)1/2 Solve 2x2 – 3x – 2 = 0 First, solve B2
– 4AC: B2 – 4AC = (-3)2
–4(2)(-2) = 9 + 16 = 25 x = - (-3)
+(25).5 = 8/4 = 2 ;
x = -(-3) – 5 = -1/2
2(2)
4 1.
x2 - 6x
+ 9 = 0 (x-3)(x-3) = 0 x = 3; x = 3 Here, we have two identical solutions
to the quadratic solution: 3, 3. 2: 2x2 - 8x + 6
= 0 2(x2 -
4x + 3) = 0 (factored
out a “2” for simplicity) (x - 1)(x
- 3) = 0 x = 1; x = 3 Another useful method for solving the
quadratic equation is completing the square: 3.
Solve x2 + 2x - 6 = 0 (x2 + 2x
+ 1) - 7 = 0 (+1, then –1 to create a perfect square of
“x + 1”) (x+1)2
= 7 x + 1 = ±√7 x = -1
±√7 4.
x2 +
2x + 6 = 0 (x2
+ 2x + 1) + 6 -1
= 0 (x + 1)2 = -5 x + 1 =
±√-5 x = -1 ± i
√5 Taking the square root of –5 resulted in
two “imaginary solutions.” 5.
The
“falling body problem”.
If s = constant: v0 t
+ ˝ at2 =
s t = time (from initial point)
˝ at2 + v0 t -
s = 0 t = -v0 ± (v02 - 4 (.5a)(-1) )1/2 2(.5) t = -v0 ± (v02 +
2a)1/2
1 |
||||||