Quadratic Equations
C.A.1-3
Solving Quadratic Equations
• If a quadratic equation ax²+bx+c=0 can’t be
solved by factoring that means the solutions
involve roots.
• The quadratic formula can be used to solve
any quadratic equation by using the
coefficients a, b, and c.
ax  bx  c  0
2
 b  b2  4  a  c
x
2a
*Since the Quadratic Formula involves the square root we can have a variety of
different solutions; from rational to radical to complex.
The derivation of the Quadratic Formula
Using the quadratic formula to solve
equations….
• Make sure your
equation is in
the form,
ax²+bx+c=0, if it
isn’t then use
algebra to put it
in this form.
• Identify the
coefficients a, b,
and c.
• Plug them into
the equation
and simplify the
result.
x  5x  3  0
2
a  1, b  5, c  3
 5  5 2  4 1  3
x
2 1
 5  25  12

2
 5  13

2

 5  13  5  13
,
2
2
These are the two
solutions, they are
radical solutions
Solve:
Examples…
x  3x  5
xx  3  5
2
x 2  3x  5  5  5
x 2  3x  5  0
a  1, b  3, c  5
 3  32  4  1  5  3  9  20  3  29
x


2 1
2
2
Examples…
Solve:
x  23  10 x
2
x  23  10 x
2
x  10 x  23  0
2
a  1, b  10, c  23
*The solutions are
approximately 6.4142 and
3.5858 if rounded to 4 decimal
places.
x
 (10) 
 102  4 1 23
2 1
10  100  92

2
10  8

2
10 2 2
 
 5 2
2
2
Solving Quadratic Inequalities
• We can use our factoring and
solution methods to
determine when a quadratic
has positive and negative
output values.
• First find the zeros for the
quadratic.
• Place them on a number line
and test values on each side
of the zeros to determine the
sign of the region.
• + means the region is positive
• - means the regions is
negative
• List all of the regions that
satisfy the inequality in
interval notation.
x  2x 1  0
zeros : x  2,1
not included
Try x=-2
Try x=0
Try x=3
+
-
+
-1
2
Solution: (-∞,-1)U(2,∞)
Examples….
Solve the
inequality:
x  3x  10  0
2
x  5x  2  0
zeros : x  5,2
not included
Try x=-3
Try x=0
+
Try x=6
-2
+
5
Solution: (-2,5)
Examples….
Solve the
inequality:
x  6x  5  0
x 1x  5  0
2
zeros : x  5,1
included
Try x=-6
Try x=-2
+
Try x=0
-5
+
-1
Solution: (-∞,-5]U[-1,∞)
Inverses-3-7 topic
p.305-313
Finding an inverse for a function…
•
•
•
•
For a function f(x)=rule, put in y=rule form.
Swap y with x and vice versa.
Solve for y.
The inverse is the function y=new rule and is
denoted:
f
1
x   new rule
Examples….
1. Find the inverse for f(x)=2x+3.
f x   2 x  3
y  2x  3
x  2y  3
Swap the x’s and y’s!
x 3  2y
x 3
y
Solve for y!
2
x 3
1
f x  
2
Examples….
x5
2. Find the inverse of f  x  
3
x5
f x  
3
x5
y
Swap the x’s and y’s!
3
y 5
x
3
3x  y  5
Solve for y!
3x  5  y
f 1  x   3 x  5