Quadratic
Equations
1
x kx k
2
Square Root Property
Example 1 & 2: Solve
1)x2 - 9 = 0.
x2  9
2) x2 + 10 = 85.
x 2  75
x   75
x 9
x  3
x   25 3  5 3
Solving Quadratic Equations by Completing the
Square: ax2 + bx + c = 0
1. Write it as : x2 + bx = c.
2. Add (b/2)2 to both sides.
3. Write it as 𝑥
𝑏 2
+
2
=𝑐+
𝑏 2
2
& Simplify
4. Solve by the square root principle.
Example 3: Solve x2 + 6x + 5 = 0 by completing the square.
x  6 x  5
2
b6
b 6
 3
2 2
2
b
2
=3
9
 
2
x 2  6 x  9  5  9
( x  3) 2  4
x3  4
x  3  2
x  3  2
x  1
x  3  2
or x  3  2
x  5
Example 4 & 5:
Solve by completing the square.
4) x 2  10 x  16  0
x2  10 x  16
x2  10 x  25  16  25
 x  5
2
 41
x  5   41
x  5  41
5) x 2  12 x  9  13
x2  12 x  36  4  36
 x  6
2
 40
x  6   40
x  6  40
x  6  2 10
Using the Quadratic Formula
To solve a quadratic equation in the standard form
ax2 + bx + c = 0, where 𝑎 ≠ 0, use the quadratic formula:
b  b 2  4ac
x
2a
There once was a Negative Boy who couldn't decide to go to
a Radical Party. Because the Boy was Square , he missed
out on 4 Awesome Chicks and it was all over by 2 AM.
Example 6: Solve x2 + 2x – 8 = 0.
Solution
a = 1, b = 2, and c = –8
2  22  4(1)(8)
x
2(1)
2  4  32
2
2  36
x
2
2  6
x
2
x
2  6
x
2
2
2  6
x
 4
2
5
Example 7: Solve x2 – 7 = 2x using the quadratic formula.
2  4  28 2  32
x

2
2
a = 1, b = –2, and c = – 7.
24 2
x
b  b 2  4ac
2
x
x2 – 2x – 7 = 0;
2a
2  (2) 2  4(1)(7)
x
2(1)
x 1 2 2
Discriminant
Nature of Solutions
b2 – 4ac = 0
We get the same solution twice.
There is one repeated solution and it
is rational.
b2 – 4ac is positive.
Two different real-number solutions
1. b2 – 4ac is a
perfect square.
The solutions are rational.
2. b2 – 4ac is not a
perfect square.
The solutions are irrational
conjugates.
b2 – 4ac is negative. There are two different imaginarynumber solutions. They are complex
conjugates.
Example 8a, b & c:
Use the discriminant to determine the number and type of solutions.
a )2 x 2  5 x  1
b) x 2  6 x  9  0
2 x2  5x  1  0
b 2  4ac 
  5   4  2 1
2
 17  0
  6   4 1 9 
2
0
1 real solution
2 real-number
solutions.
c)0.3 x 2  0.4 x  0.8  0
  0.4   4  0.3 0.8 
2
 0.16  0.96
 0.8
No real solutions
(2 Imaginary or non-real
complex solutions)
8
y - intercept
y
6
5
4
3
f (x) = ax2 + bx + c
2
x - intercepts
1
-5 -4 -3 -2 -1
-1
-2
1
2
3
4
5
x
-3
-4
-5
 b
vertex :   , f
 2a
2

b
b
4
ac

b



 
.
   or   ,

4a 
 2a  
 2a
Example 9: Write f(x) = x2 – 6x + 8 in the form f(x) = a(x – h)2 + k. Then determine
whether the graph opens upwards or downwards, find the vertex and axis of symmetry,
and draw the graph. Find the domain and range.
•Solution Complete the square.
x-intercept: Set f(x) = y = x2 – 6x + 8
y = x2 – 6x + 8
(x – 2)(x – 4) = 0, so x =2 & 4
a = 1 > 0 up
y – 8 = x2 – 6x
V (3, 1)
y – 8 + 9 = x2 – 6x + 9
y + 1 = (x – 3)2
Axis: x = 3
2
y = (x – 3) – 1
•
x
y
(x, y)
0
2
3
4
6
8
0
–1
0
8
(0, 8)
(2, 0)
(3, –1)
(4, 0)
(6, 8)
y-intercept: set x = 0
x-intercept
vertex
x-intercept
Domain: (−∞,∞ ),
Range: [−1,∞ )
10
Example 10:
A farmer has 200 ft of fence with which to form a rectangular pen on his
farm. If an existing fence forms one side of the rectangle, what
dimensions will maximize the size of the area?
Solution
w
l
w
A = (200 – 2w)w
Existing fence
= 200w – 2w2
2w  l  200
Alw
b
w
 50
2a
The maximum area,
5000 ft2, occurs
when
w = 50
ft and l = 200 –
2(50), or 100 ft.