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College Algebra
Summary 4
Quadratic Equations.
Definition: a quadratic equation is any equation that can be written in the form
2
ax  bx  c  0 where a, b, and c are real numbers and a≠0.
2
ax  bx  c  0 is called the standard form of a quadratic equation. It is also called a
second-degree equation because the left side is a polynomial of degree 2.
According to the Fundamental Theorem of Algebra, a polynomial equation of degree n can have
as many as n solutions. Therefore, in a quadratic equation you expect to get as many as 2
different solutions.
Definition of
2
Definition: The algebraic expression b  4ac , is call the DISCRIMINANT of the quadratic
equation.
Nature of the roots or solutions of a quadratic equation with integer coefficients a, b, c.
2
Two complex roots
b  4ac <0
2
One rational root. The root is said to have
b  4ac =0
multiplicity two (double root)
2
Two real rational roots
b  4ac >0 and a perfect square
2
b  4ac >0 and not a perfect square Two real irrational roots
If a, b, and c are real numbers and the discriminant b 2  4ac of the quadratic equation
ax 2  bx  c  0 is negative, then the two solutions of the equation are complex. This is an
indication that the quadratic function f ( x)  ax 2  bx  c has no real zeros or x-intercepts.
Example 1) x 2  3x  0 is a quadratic equation where a = 1, b = -3 and c = 0
The discriminant is b 2  4ac  (3) 2  4(1)( 0)  9  0  9
The two solutions of the equation are equal and it is a rational number
Example 2) 5x 2  3x  4  2x 2  5x  2 is equivalent to 7x 2  8x  6  0 , therefore it is a
quadratic equation where a = 7, b = -8 and c = 6.
The discriminant is b 2  4ac  (8) 2  4(7)(6)  64  168  104
The two solutions of the equation are complex conjugate
Example 3) 7x 2  8x  1  0 , therefore it is a quadratic equation where a = 7, b = -8 and c = 1.
The discriminant is b 2  4ac  ( 8) 2  4(7)(1)  64  28  36  6 2
The two solutions of the equation are different and rational numbers
-1-
1
2

 5 is equivalent to 1( x  1)  2( x  3)  5( x  3)( x  1),
x  3 x 1
x  3, x  1
Example 4)
 x  1  2x  6  5x 2  10x  15, x  3, x  1  5x 2  7 x  20  0, x  3, x  1.
a quadratic equation where a = 5, b = 7 and c = -20.
The discriminant is b 2  4ac  (7) 2  4(5)( 20)  49  400  449
The two solutions of the equation are different and irrational numbers
How to solve a quadratic equation?
The following methods or techniques are available for solving a quadratic equation
a) Factoring method
2
b) Square root method: Note: If ax  b   k then ax  b   k
b) Completing the square method
c) Using the quadratic formula x 




 b  b 2  4ac
2a
I. Factoring method. This method can only be used when the discriminant of the
equation is a perfect square.
Steps to be followed if the discriminant is a perfect square
2
Step 1. Write the quadratic equation in the standard form ax  bx  c  0
Step 2. Write the quadratic equation in factored form
Step 3. Set each factor equal to zero
Step 4. Solve the linear equation that result when each factor is set to zero. The solutions are
obtained.
Examples:
1) Solve 2x 2  3x  2  0
Note : b 2  4ac  ( 3) 2  4( 2)( 2)  9  8  1, which is a perfect square ;
2x 2  3x  2  0  ( 2x  1)( x  2)  0  2x  1  0 or x  2  0
1
or x  2
2
2) Solve 5x 2  3x  2  x 2  20x  4
x
The equation is equivalent to 4x 2  23x  6  0
Note : b 2  4ac  ( 23) 2  4( 4)( 6)  529  96  625  25 2 , which is a perfect square ;
4x 2  23x  6  0  ( 4x  1)( x  6)  0  4x  1  0 or x  6  0
x
1
or x  6
4
-2-
1
6

 1 is equivalent to 1( x  1)  6( x  3)  1( x  3)( x  1),
x  3 x 1
x  3, x  1
3) Solve
 x  1  6x  18   x 2  2x  3, x  3, x  1  x 2  9x  14  0, x  3, x  1.
x 2  9x  14  0  (x  2)( x  7)  0  x  2  0 or x  7  0  x  2 or x  7
The solution set of the equation is {-2, -7}
4) Solve
x2  4
Solve the equation
7
x2
x2  4
x2  4
7 
( x  2)  7 ( x  2)  x 2  4  7x  14  x 2  7x  10  0
x2
x2
 ( x  2)( x  5)  0  x  2 or x  5
but x  2 because the denominator of a fraction can not be zero.
Therefore the solution set is {5}
II. Square root method: Note: If
resulting linear equations.
ax  b 2  k then ax  b   k . Solve the
Example 1 : solve the equation ( 2x  5) 2  49
Solution : ( 2x  5) 2  49  2x  5   49  2x  5  7
5 7
12
2
x
or x 
 x  6 or x  1
2
2
2
The solution set of the equation is {-1 , 6}
 2x  5  7  x 
Example 2 : solve the equation (3x  2) 2  7
Solution : (3x  2) 2  7  3x  2   7  3x  2   7
 3x  2  7  x 
2 7
2 7
2 7
x
or x 
3
3
3
Example 3 : solve the equation 4x 2  12x  9  25
Solution : 4x 2  12x  9  25  ( 2x  3) 2  25  2x  3   25  2x  3  5
 2x  3  5  x 
3 5
3 5
35
 x
or x 
 x  1 or x  4
2
2
2
-3-
III. Completing the square method. This technique can be used always, regardless of the
nature of the discriminant.
Steps to be followed:
2

Step 1. Write the quadratic equation in the standard form ax  bx  c  0

Step 2. Subtract the number c from both members of the equation to obtain
ax 2  bx  c

Step 3. Divide both member of the new equation by the number “a” to obtain
x2 
b
c
x
a
a
2


b
b2
1 b
, that is , find    
a
2 a
4a 2
Step 5. Add the result of step 4 to both members of the equation in step 3. This is called
completing the square.
Step 4. Find the square of half of
Obtain:

b
b2
c b2
x  x
 
a
a 4a 2
4a 2
2
Step 6. Factor the left member of the equation and combine the terms of the right member
to obtain
2
b 
b 2  4ac

x  
2a 

4a 2

Step 7. Use the square root method to obtain
b
b 2  4ac

x   
2a 

4a 2

Step 8. Solve for x
b
b 2  4ac
x

2a
4a 2
 b  b 2  4ac
x
2a
This is called the quadratic formula
-4-
Example 1:
Use completing the square to solve the equation 3x 2  6x  5  x  x 2
Step 1. 3x 2  6x  5  x  x 2 is equivalent to 2x 2  5x  5  0
Step 2. 2x 2  5x  5
Step 3. x 2 
5
5
x
2
2
2
2
1 b
1 5

Step 4.  .         
2 a
 2 2

5
25 5 25
Step 5. x 2  x 
 
2
16 2 16
2
5
25
 
4
16
2
5
65

Step 6.  x   
4
16

Step 7.
x
5
65

4
16
Step 8.
x
5
65
5  65
5  65
5  65

x
x
or x 
4
4
4
4
4
IV. Using the quadratic formula
 b  b 2  4ac
x
2a
which was obtained by
following the 8 steps in completing the square.
Example 1:
Use the formula to solve the equation 2x 2  5x  5  0
Solution : a2, b  5, c  5 
 b  b 2  4ac  ( 5) 
x

2a
x
 52  4( 2)( 5)
2( 2)
5  25  40 5  65
5  65
5  65

. That is x 
or x 
4
4
4
4
The two solutions are irrational.
-5-
Example 2 : Use the formula to solve the equation x 2  4x  13  0
Solution : a1, b  4, c  13 
 b  b 2  4ac  ( 4) 
x

2a
 42  4(1)(13)
2(1)
4  16  52 4   36

2
2
4  6i
4  6i
x
or x 
. That is x  2  3i or x  2  3i
2
2
x
The two solutions are complex numbers
Example 3
The quadratic equation x 2  4x  29  0 has discriminant equal to
b 2  4ac  ( 4) 2  4(1)( 29)  16  116  100
The two solutions are complex:
 b  b 2  4ac  ( 4)   100 4  10i
x


 2  5i
2a
2(1)
2
The solutions are 2 + 5i and 2 – 5i
Quadratic equations conclusion.
A quadratic equation is an equation of the form ax 2  bx  c  0 where a, b, c are numbers and
a  0.
The solutions of a quadratic equation can be obtained by using the quadratic formula
 b  b 2  4ac
2a
Example. Find the solution of the equation 2 x 2  x  6  0
x
 b  b 2  4ac 1  1  4(2)( 6) 1  49 1  7
3
a=2, b=-1, c=-6, x 



  or 2
2a
2(2)
4
4
2
Example 1: Solve the following quadratic equation with absolute value.
| x 2  8 | 12
Solution : | x 2  8 | 12  x 2  8  12 or x 2  8  12
 x 2  4 or x 2  20
 x   4 or x    20
 x  2 or x   4( 1)(5)  x  2 or x  2i 5
The solutions are :  2, 2, 2i 5 and  2i 5
-6-
Example 2: Solve the following quadratic equation with absolute value
x 2  x  12  x 2  x  12 or x 2  x  12
x 2  x  12  x 2  x  12  0  x  4)( x  3)  0  x  4 or x  3
 1  1  4(1)(12)  1   47

2
2
 1  i 47
1
47
1
47
x
 x i
or   i
2
2
2
2
2
x 2  x  12  x 2  x  12  0  x 

1
47
1
47 
The equation has 4 solutions :   4, 3,   i
,  i

2
2
2
2 

Example 3. Solve | x 2  3x | 0
Solution : | x 2  3x | 0  x 2  3x  0
 x( x  3)  0
 x  0 or x  3  0
 x  0 or x  3
The solutions are : 0,  3
Properties of quadratic equations
a) If x1 and x 2 are the roots of the quadratic equation ax 2  bx  c  0
b
a
b) If x1 and x 2 are the roots of the quadratic equation ax 2  bx  c  0
then the sum of the roots x1  x 2  
c
a
c) If x1 and x 2 are the roots of a quadratic equation then the
then the product of the roots x1 x 2 
equation is given by x 2  x1  x 2 x  x1x 2  0
Problem 1: without solving the equation 2x 2  3x  5  0 , find the sum and the product of the
two solutions.
a = 2, b = -3, c= -5
b
3 3
c 5
5
sum x1  x 2    
 , product x1x 2  

a
2
2
a
2
2
-7-
Problem 2: find the quadratic equation with roots -2 and 7.
sum x1  x 2  2  7  5, product x1x 2  (2)(7)  14
The equation is given by x 2  x1  x 2 x  x1x 2  0
The equation is x 2  5x  14  0
Problem 3: find the quadratic equation with roots 7 and -10 .
sum x1  x 2  7  10  3 product x1x 2  (7)( 10)  70
The equation is given by x 2  x1  x 2 x  x1x 2  0
The equation is x 2  ( 3)x  ( 70)  0
The equation is x 2  3x  70  0
Problem 4: find the quadratic equation with roots 
3 1
6
5
1
sum x1  x 2      

5 2
10 10
10
3
1
and
5
2
3
 3  1 
product x1x 2       
10
 5  2 
The equation is given by x 2  x1  x 2 x  x1x 2  0
 1
  3
The equation is x 2     x  
0
 10 
 10 
1
3
The equation is x 2  x 
0
10
10
The equation is 10x 2  x  3  0
-8-