Transcript Algebra I- Module 4 Solve Quad Alg Honors Topics

Honors Topics




You learned how to factor the difference of two
perfect squares: x 2  y 2  ( x  y )( x  y )
2
Example: x  1  ( x  1)( x  1)
But what if the quadratic is x 2  1 ? You learned that it
was not factorable. It is factorable, but with complex
factors.
How does it work? Let’s solve an equation.
If you solve the equation x 2  1  0, you get solutions
x =1,-1. If you solve the equation x 2  1  0, you get
the solutions x = i ,-i.
Since a quadratic factors into (x – zero)(x-zero):
x 2  1  ( x  i)( x  i)

Factor each quadratic expression.
x 9 
2
 x  3i  x  3i 
x  100 
( x  10i)( x  10i )
2
x  64 
( x  8i )( x  8i )
2
4
x 

49
2 
2 

 x  i  x  i 
7 
7 

2

You must always simplify powers of the
imaginary unit, i. You learned the properties:
i² = -1 and 1  i
But how would you simplify i ?
You need to find out how many pairs/sets of
imaginary units, i, you have and simplify them.
8
Since every i² = -1, i  i  i  i  i  i  i  i  i  i
8

2 4
 (1)4  1
Simplify each imaginary number.

i  i
28
i  i
 (1)  1
2 39

 i 
 (1)  1
2 7
 (1)  1
i  i
14
i
i
 (1)  1
14

44
78

2 14
even
2 22
22
39
7
 1 or  1
i  i
17
i
49

 i 
i
83
 i  (1)8  i  1 i  i
2 24
i  i
35
2 8

 i 
i
 i  (1)24  i  1 i  i
2 17
 i  (1)17  i  1 i  i
2 41
 i  (1)41  i  1 i  i
odd
 i or  i

Recall that you must rationalize the denominator
of a fraction. You cannot leave a radical on the
bottom of a fraction. You also cannot leave a
complex number on the bottom of a fraction.
Divide by a Complex Number a + bi
Multiply numerator and denominator by
the conjugate of the bottom, a – bi
Conjugates:
4-6i, 4+6i

½ + i, ½ - I
7i, -7i

Divide.
1

2i
1 (2  i )

 2  i  (2  i)
3i

1  5i
3i (1  5i )

1  5i  (1  5i)
3i  15i
3i  15( 1)
2i
2i




2
2
1  25i
1  25(1)
4i
4  (1)
15  3i
15 3
2i
2 1
or
 i
or  i
26
26 26
5
5 5
2
2  3i

4  2i
(2  3i ) (4  2i )


(4  2i ) (4  2i )
8  4i  12i  6i 2

2
16  4i
8  16i  6( 1)

16  4( 1)
8  16i  6 2  16i


16  4
20
2(1  8i ) 1  8i 1 4

  i
20
10
10 5


Solve a quadratic equation in standard form
2
with leading coefficient of 1 x  bx  c  0
x 2  3x  2  0
( x  2)( x  1)  0
x  2  0 or x  1  0
x  2 or x  1
Factored form is (x-m)(x-n) = 0
Therefore x = m, x = n are the roots.
m + n = -2 + -1 = -3, which is -b
m · n = (-2)(-1) = 2, which is c
If a = 1 and x 2  bx  c  0, then:
Sum of the roots: m + n = -b
Product of the roots: m · n = c
2
x
 (m  n) x  (m  n)  0
or
Note: if a≠1, sum = -b/a and product = c/a







Given the zeros of the quadratic x 2  bx  c  0,
write an equation in standard form.
x = 5, -4
{-6, 0}
x = 5±3i
5+-4 = 1
So -b = 1
b = -1
5(-4) = -20
So c = -20
x 2  x  20  0
-6+0 = -6
So –b = -6
b=6
(-6)(0) = 0
So c = 0
x2  6x  0
(5+3i)+(5-3i) = 10
So –b = 10
b = -10
(5+3i)(5-3i)= 34
So c = 34
x 2  10 x  34  0



Recall the standard form of quadratic
2
ax
 bx  c  0 .
equation:
The discriminant, b2  4ac , of the quadratic
indicates the nature of the roots.
2 real roots
1 real root
2 complex roots
b 2  4ac  0


b 2  4ac  0
b 2  4ac  0
Q: How would you find the value of a, b, or c
in a quadratic equation given that there are 2
real roots? 1 real root? 2 complex roots?
A: Set up an equation or inequality and solve!


Given x 2  2 x  c  0, find the value(s) of c if there
is/are:
2 real roots
1 real root
2 complex roots
b 2  4ac  0
b 2  4ac  0
b 2  4ac  0
2  4(1)c  0
4  4c  0
4c  4
c 1
22  4(1)c  0
4  4c  0
4c  4
c 1
22  4(1)c  0
4  4c  0
4c  4
c 1
2
*C is less than 1
*C is equal to1
*C is greater than 1

You learned how to solve quadratic equations using different
methods and you learned how to solve quadratic inequalities
by graphing. Now let’s look at solving inequalities
algebraically.
Solving Quadratic Inequalities Using Algebra
1. Write the inequality as an equation ax 2  bx  c  0.
2. Solve the quadratic equation by factoring.
3. Graph each solution on the same number line.
4. Test a number in all three sections of the number line by
substituting it into the original inequality. If the inequality is
True, each number in that section is a solution.
5. Write your solutions as an inequality. If m and n are the
zeros for the quadratic equation, the solutions will be:
solution 1: m < x < n
OR
solution 2: x< m or x > n






Solve the inequality
x2  x  2  0
Write it as an equation and find the roots
Graph the roots -2, 2 on the number line
Use a closed circle since it is ≤
Check values in each section
x = -2, 0, and 3
(2) 2  (2)  2  0
422 0
False, -2 is not shaded
(0) 2  (0)  2  0
002 0
True, 0 is shaded
x2  x  2  0
( x  2)( x  1)  0
x  2  0 or x  1  0
x  2 or  1
(3) 2  (3)  2  0
93 2  0
False, 3 is not shaded

Solution: -1 ≤ x ≤ 2

Note: if the inequality is x  x  2  0, then the False sections
would now be true. The solution would be x < -1 or x > 2.
2