Download MA120 Worksheet3B

MA120 Worksheet3B
This worksheet continues with the topic of quadratic equations. Here we derive the quadratic formula and then apply this formula
to solve the various types of quadratic equations. Finally, we also show some problems modelled by quadratic eqautions.
Sec 1.5 Quadratic Equations (continued)
Deriving the quadratic formula
We begin with the generic formula for a quadratic equation:
ax 2  bx  c  0
x2 
Divide by the coefficient
2
 b 
 to both sides of the equation to complete the square
 2a 
b
c
x 0
a
a
Add 
2
b
c
 b 
 b 
x  x    0 
a
a
 2a 
 2a 
2
2
2
a , a  0
Write the completed square and move all other tems on the right
2
b 
c

 b 
x 
   
2a 
a

 2a 
Open the square on the right side
2
b 
b2
c

x 
  2 
2a 
a
4a

Rewrite the right side as a single fraction using 4a as LCD
b 
b 2  4ac

x
 
2a 
4a 2

Now solve for
2
2
x using the definition of a square root :
if u  d then
2
x
b
b  4ac

2a
4a 2
2
Remove 4a from the radical and move
b
b 2  4ac
x

2a
2a
x1, 2 
u d
2
b
to the right side
2a
Rewrite the right side as a single fraction, using
 b  b 2  4ac
2a
2a as LCD
This is the quadratic formula for finding the root(s) of a quadratic
equation,
x1 and x 2 , if they exist.
Note on the Discriminant
The expression b  4ac is called the Discriminant, as it determines (“discriminates”) the number of roots (or solutions) that
2
the quadratic equation has. If the Discriminant b  4ac > 0 , then it has two real roots, and adding and subtracting
2
b 2  4ac from  b obtains two solutions for the equation. If the Discriminant b 2  4ac  0 , then b 2  4ac  0 ,
and that gives only one solution to the equation.
1
If the Discriminant b  4ac < 0 , then it has no real number roots and the equation also has no real number solutions. In
this case the roots of the Discriminant are two complex numbers of the type z  bi . The coefficient b is a real number,
2
and i is an imaginary number that is a solution to the equation x  1 . Adding
two complex solutions for the equation.
2
 bi to  b in the quadratic formula gives
Example1
A quadratic equation with two real solutions:
2x 2  6x  1  0
 b  b 2  4ac
2a
 (6)  36  8 6  28


4
4
6 2 7 3 7


4
2
In this case
a  2 , b  6 , c  1
x1, 2 
Write the formula and substitute
x1, 2
Simplify
x1, 2
3  7 3  7 
,

2 
 2
The two real solutions are: 
Solve:
2x 2  2x  1  0
2 x 2  3x  4  0
Example2
A quadratic equation with one real solution:
9x 2  6x  1  0
 b  b  4ac
2a
 (6)  36  36 6  0
x1, 2 

18
4
3
x
2
x1, 2 
In this case
a  9 , b  6 , c  1
2
Write the formula and substitute
Simplify
3
2
The real solutions is:  
Solve:
x 2  4x  4  0
x 2  2x  1  0
2
Example3
A quadratic equation with two complex solutions:
x2  2x  2  0
x1, 2 
 b  b 2  4ac
2a
 (2)  4  8 2   4

2
2
2  2i

 1 i
2
x1, 2 
x1, 2
In this case
a  1 , b  2 , c  2
Write the formula and substitute
Note:

 4   1  ( 4 )  i   2  2i
Solve:
3x 2  2 x  4  0
3x 2  8 x  7  0
Applications
A persons normal systolic blood pressure is measured in millimeters of mercury (mm Hg) and is modeled by the following
quadratic equation:
P  0.006 A 2  0.02 A  120 where A is age. Find the age for a man whose normal systolic pressure is 125 mm Hg.
Solution: In this case we substitute in the formula for P and solve the quadratic equation for A .
The equation has two solutions: A  31 and A  27 . We consider the solution A  31.
Solve:
Using the quadratic equation above find the age of a person whose normal systolic pressure 115 mmHg.
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Short Quiz3
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