# Algebra II Chapter 5 Notes and Practice- Solving Quadratic Equations

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```					           Algebra II Chapter 5 Notes and Practice – Solving Quadratic Equations

Recall that a quadratic equation can take one of two forms. Those forms are Standard
form, which is, y  ax 2  bx  c or vertex form, which is y  a ( x  h) 2  k . We are now going to
solve quadratic equations. If a quadratic equation starts in standard form and “ b  0 ” we will
use two different methods to solve it: Factoring and The Quadratic Formula. If the quadratic
equation is in vertex form or if "b  0" we will solve it using the square root method. Your
answers can be Real Numbers or they could be Imaginary Numbers.

Solving a Quadratic Using the Square Root Principle.

Solving a quadratic of the form a( x  h)2  k  0 or ax 2  c  0

To solve a quadratic equation in vertex form or to solve a quadratic whose “b” value = 0
we can use the square root principle.

How to Use the Square Root Method to solve a quadratic

2.       Take the square root of each side of the equation
x2  c
Recall that if x 2  c and “c” is a positive number then
so x   c

3.       If, after taking the square root of the quadratic expression the “x” is
still not isolated continue solving until it is.

Examples of Solving a Quadratic Equation Using Square Root Method.

Example 1 Solve: x 2  20  0                             Example 2. Solve: 3x 2  27  0

Solution :
Solution :
Step1: 3 x 2  27  27  0  27
Step1: x  20  20  0  20
2
3 x 2 27
x  10
2                                                             
3      3
Step 2 :              x 2  20                                                   x  9
2

x   20                             Step 2 :             x 2  9
x   45                                                  x  i 9  3i
x  2 5
Example 3: Solve: 4( x  3) 2  6  18          Example 4:        Solve:       2( x  5) 2  4  20

Solution:                                       Solution:

2( x  5) 2  4  4  20  4
4( x  3) 2  6  6  18  6                              2( x  5) 2 24

4( x  3) 2 12                                       2          2
                                        ( x  5)  12
2
4       4
( x  3)  3
2
( x  5) 2  12
( x  3) 2  3                                     x  5  i 4  3
x3  3                                        x  5  2i 3
x  3  3  3  3                              x  5  5  5  2i 3
x  3  3                                       x  5  2i 3

Solving Quadratics in Standard form Using Factoring

If a quadratic equation starts in Standard form that is ax 2  bx  c =0 form then one
method for solving these is factoring. Note: The quadratic must be equal to zero before
checking for factorability. First you will need to check to see if the quadratic is factorable by
multiplying “a x c” and seeing if there are factors of that product that add to “b”. If there are
then you know the quadratic is factorable and you can use the process below to solve.

2.        Find the “zero” of each factor. What ever the zeros are represent the
solutions to the equation.

Solve the following using factoring

Ex 1. x 2  5 x  14  0                        Ex: 2 4 x 2  12 x  9  0

Step 1 Factor                                   Step 1 factoring

( x  7)( x  2)  0                            4 x2  6 x  6 x  9  0
2 x( x  3)  3(2 x  3)  0
Step 2 Find the zeros of each factor            (2 x  3)(2 x  3)  0

x  7, x  2                                   Step 2 Find the zero

2x  3  0
3
x     double root
2

Unfortunately not all quadratics in standard form are factorable so we need to have an
additional method to solve these. This method requires you to use the Quadratic Formula. The
quadratic formula is a formula that you substitute the “a” , “b”, and “c” values into to find the
two solutions.

b  b 2  4ac
2a

Part of this formula contains a value underneath the radical “ b2  4ac ”. This value underneath
the radical is called the discriminant of the formula. The value of the discriminant will tell you
how many solutions you have and what type of solutions they are.

Determining What types of solutions you have from the discriminant.

Discriminant > 0 (Positive)

If the value of the discriminant "b2  4ac  0" This tells you the equation has two real
solutions.

Discriminant = 0

If the value of the discriminant "b2  4ac  0" This tells you the equation has one real
b
solution and the solution =     . Note: Quadratic is factorable if discriminant = 0.
2a

Discriminant < 0 (Negative)

If the value of the discriminant "b2  4ac  0" This tells you the equation has no real
solutions but it does have two complex conjugate solutions. Note: IF you are asked to solve
over the set of real numbers only and the discriminant is negative, then you would state
there are no real solutions and you would be finished.

1.      Identify the “a”, “b”, and “c” value from the quadratic equation. Make sure equation
= 0.
2.      Find the value of the discriminant to determine what types of solutions you have.
Note: If solving over the set of Real numbers and the discriminant is negative
then you would state there are no real solutions and you would be finished. If
you are solving for solutions over the set of complex numbers you would
continue to step three.
b
3.       If the discriminant is zero find the solution by determining the value of    . If the
2a
discriminant is not zero then plug the value of the discriminant and the other values
into the quadratic formula and solve for the solutions.
4.       Make sure the final answer is simplified

Examples:

Solve over the set of complex numbers

Ex 1.   2 x2  4 x  3  0

Step 1: Identify a, b, and c values
a = 2, b = -4, and c=-3

Step 2: Find the value of the discriminant " b2  4ac "
(4)2  4(2)(3)
16  24  40 (Two real solutions)
Step 3: Since the discriminant is not zero plug it in with other values into quadratic
formula.

(4)  40 4  2 10 2(2  10)
x                        
2(2)        4         4
2  10
x
2
Ex 2:   x2  5x  7  0

Step 1: a = 1, b= 5, c=7

(5)2  4(1)(7)
Step 2:
25  28  3 (two complex conjugate solutions)

(5)  3
x
2(1)
Step 3:
5  i 3
x
2

Solve the following Problems using the Square root principle.

1. 3( x  3) 2  3  0      2. 2( x  4) 2  10        3. 4( x  1) 2  5  5

Solve the following Problems using Factoring

4. x2  7 x  8  0         5. 5 x 2  5  0            6. 4 x2  x  5  0

Solve the following Problems using the quadratic formula over the set of complex numbers

7. 3x 2  5 x  1  0       8. 2 x 2  4 x  1  0      9. 2 x 2  4 x  5

Find the value of the discriminant for the given quadratic. From the value of the
discriminant determine how many solutions there are and the type of solutions.

10. 5 x 2  3x  4  0     11. 6 x 2  2 x  1  0     12. x 2  2 x  1  0

Discriminant=________       Discriminant=________       Discriminant=________

# of Solutions = ________   # of Solutions = ________   # of Solutions = ________

Solve using any method over the set of real numbers

13. 4 x 2  9  0           14.    x 2  4 x  12  0   15. 2 x2  4 x  3  0

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