College Algebra Lecture Notes
Section 1.5
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Section 1.5: Solving Quadratic Equations
Big Idea: Quadratic equations can be solved under any circumstance if we allow complexvalued solutions.
Big Skill: You should be able to solve any quadratic equation by factoring and using the zero
product property, completing the square and using the square root property of equality, and by
using the quadratic formula.
A. Solve Quadratic equations Using the Zero Product Property
 A quadratic equation is an equation in the variable x with constants a, b, and c
(sometimes called parameters) that can be written in the form ax 2  bx  c  0 .
 The standard form for a quadratic equation is as above, where all non-zero terms are on
the left hand side, and those terms are written in descending powers of the variable.
Quadratic Equations
A quadratic equation can be written in the form ax 2  bx  c  0 with a, b, c  and a  0 .
Zero Product Property
If A and B represent real numbers or real-valued expressions,
and A  B  0 ,
then A  0 or B  0 .

To solve a quadratic equation using the zero product property:
o Get all terms on one side of the equation (which means zero will be on the other
side)
o Factor the non-zero side
o Set each factor equal to zero (justified by the zero product property)
o Solve each mini “factor equation.”
Practice:
1. Solve: 2 x 2  7 x
College Algebra Lecture Notes
Section 1.5
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2. Solve: 5 x  2 x 2  3
3. Solve: 8 x 2  24 x  18
B. Solve Quadratic equations Using the Square Root Property of Equality
Square Root Property of Equality
If X represents an algebraic expression,
and X 2  k ,
then X  k or X   k
which is also written as X   k .

To solve a quadratic equation using the square root property of equality:
o Manipulate the equation until the left hand side is a perfect square, and there is a
single number on the right hand side.
o Apply the square root property of equality.
College Algebra Lecture Notes
Practice:
4. Solve: 4 x 2  12  100
5. Solve: x 2  12  4
6. Solve:  x  7   5  25
2
Section 1.5
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College Algebra Lecture Notes
Section 1.5
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C. Solve Quadratic Equations by Completing the Square
To Solve a Quadratic Equation by Completing the Square:
(i.e, writing a quadratic trinomial as a perfect square trinomial plus a constant)
 Get the constant term on the right hand side of the equation.
i.e., if x 2  bx  c  0 , then write the equation as x 2  bx  c

Make sure the coefficient of the square term is 1.

Identify the coefficient of the linear term; multiply it by ½ and square the result.
1 
i.e., Find the number b in x 2  bx  c and compute  b 
2 

2
Add that number to both sides of the equation.
2
1 
1 
i.e., x  bx   b   c   b 
2 
2 
2
2

Write the resulting perfect square trinomial as the square of the binomial .
2
1 

1 
i.e.,  x  b   c   b 
2 

2 

2
Use the square root property to solve the equation.
Practice:
7. n 2  10n  6  0
College Algebra Lecture Notes
8. x 2  16 x  7  0
9. z 2  7 z  1  0
10. 3 x 2  6 x  5  0
Section 1.5
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College Algebra Lecture Notes
Section 1.5
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D. Solve Quadratic Equations Using the Quadratic Equation
The Quadratic Formula
The solution(s) to the quadratic formula ax 2  bx  c  0 (for a  0) are given by the quadratic
formula:
x
b  b 2  4ac
2a
Proof:
The quadratic formula is derived by completing the square on the standard from of a quadratic
equation:
 Get the constant term on the right hand side of the equation.
ax 2  bx  c  0


ax 2  bx  c
Make sure the coefficient of the square term is 1.
ax 2  bx  c
ax 2  bx
c

a
a
b
c
x2  x  
a
a
Identify the coefficient of the linear term; multiply it by ½ and square the result.
2

b2
1 b




4a 2
2 a
Add that number to both sides of the equation.
b
c
x2  x  
a
a
2
b
b
c b2
2
x  x 2   2
a
4a
a 4a
2
b
b
b2 c
x2  x  2  2 
a
4a
4a a
College Algebra Lecture Notes

Section 1.5
Write the resulting perfect square trinomial as the square of the binomial .
b
b2
b2 c
x2  x  2  2 
a
4a
4a a
2
b 
b 2 4ac

x





2a 
4a 2 4a 2

b  b 2  4ac

x  
2a 
4a 2

Use the square root property to solve the equation.
2

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b 
b 2  4ac

x





2a 
4a 2

2
x
b
b 2  4ac

2a
2a
x
b
b 2  4ac

2a
2a
b  b 2  4ac
x
2a
College Algebra Lecture Notes
Section 1.5
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E. Use the Discriminant to Identify Solutions
 b2 – 4ac is called the discriminant. The discriminant is important because it determines
the nature of the solutions (roots) of the quadratic equation.
Examples of the Nature of the Roots of a Quadratic Equation with Rational Coefficients:
1. If b2 – 4ac is positive and a perfect square, then there are two solutions that are real,
rational, and unequal. In this case, you can also solve the quadratic equation by
factoring.
Example: 2 x 2  5 x  1.125  0
2. If b2 – 4ac is positive but not a perfect square, then there are two solutions that are
real, irrational, and unequal.
Example: 1.5 x 2  7 x  3  0
3. If b2 – 4ac = 0, then there is just one solution (a repeated root) that is real and
rational (or we can say that the two solutions are equal). This case can also be solved
by factoring.
Example: 9 x 2  24 x  16  0
College Algebra Lecture Notes
Section 1.5
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4. If b2 – 4ac is negative, then there are two solutions that are complex and unequal.
Example: 4 x 2  2 x  9  0
Practice:
5. Solve for k: 16k 
9
 24
k
F. Solve Applications of Quadratic Equations
Practice:
6. The length of a tennis court is 12.8 m more than its width. If the area of a tennis court is
262 m2, what are its dimensions?
College Algebra Lecture Notes
Section 1.5
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7. The golden ratio is important in architecture and design because it is the foundation for
the most pleasing looking rectangles and linear proportions. The golden ratio is the ratio
of the length to the width of a rectangle such that when you remove a square that has side
equal to the width of the rectangle, the remaining rectangle has sides that also are in
proportion to the golden ratio. Calculate the golden ratio.