Download 3.4 Solving Quad Equations with the Quad Formula

Chapter 3 Quadratic Equations
3.4
Solve the following Quadratic equations:
x2 = 49
2x2 – 8 = 40
Solve by completing the square.
x2 – 7x = - 11
3x2 + 18x – 2 =0
Solve the General Form of the equation for x by
completing the square.
ax2
+ bx + c = 0
 2 b
b 2  ___
b 2   c  0
a x  x  ___
a

4a 2 4a 2 
 2 b
b2 
b2
a x  x  2   c 
0
a
4a 
4a

b  4ac  b

a x   
0
2a 
4a

2
b 
4ac  b 2

a x    
2a 
4a

2
2
2
b
b
 4ac


x  
2
2a 
4a

2
2
b
b
 4ac


x   
2
2
a
4
a


b
b 2  4ac
x

2a
2a
Draw a Graphic Organizer of the concepts we have learned so far.
The Quadratic Formula
The solution of the quadratic equation
ax2 + bx + c = 0 can be found by using the
quadratic formula:
b  b  4ac
x
2a
2
Solving Quadratic Equations Using the Quadratic Formula
Solve 2x2 - 5x + 2 = 0.
a = 2, b = -5, c = 2
b  b 2  4ac
x
2a
(5)  (5)  4(2)(2)
x
2(2)
2
5  25  16
x
4
5 9
x
4
53
x
4
8
x
4
x=2
or
2
x
4
1
x
2
Solving Quadratic Equations Using the Quadratic Formula
Solve x2 - 6x + 7 = 0.
b  b 2  4ac
x
2a
(6)  (6)2  4(1)(7)
x
2(1)
6 8
x
2
62 2
x
2
x 3 2
x 3 2
x 3 2
Solving Quadratic Equations With No Real Roots
Solve x2 - 5x + 7 = 0.
b  b 2  4ac
x
2a
(5)  (5)  4(1)(7)
x
2(1)
2
5  3
x
2
Since 3
is not defined by
real numbers, then
this equation has
NO REAL ROOTS.
Solving Quadratic Equations With No Real Roots Using Complex Numbers
An equation such as x2 + 1 = 0 (x2 = -1) has no solution in
the set of real numbers. But, by extending the number
system, we can give meaning to the solution of this
equation. We do this by defining i with the property that:
i2 = -1 or i = √ - 1
Since there is no real number that has its square
as a negative, the number i is not a real number.
It cannot be expressed as a decimal and it can
not be expressed as a point on the number line.
For these reasons, the square roots of negative
numbers are called imaginary numbers.
Solving Quadratic Equations With No Real Roots
Solve x2 - 6x + 13 = 0.
(6)  (6)2  4(1)(13)
x
2(1)
6  16
x
2
6  4i
x
2
1  i
16  16  1
= 4i
x  3  2i
The roots of the equation are
x = 3 + 2i and x = 3 - 2i.