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Intermediate Algebra
Test #7 Review
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Sections 8.1: Solving Quadratic Equations by Completing the Square
The Square Root Property
If x 2  p then x  p or x   p .
To Solve a Quadratic Equation Containing Only a Square and Constant Term:

Isolate the square term.

Use the square root property.
Comparison Example:
Solve x 2  16  0 using factoring.
x 2  16  0
 x  4  x  4   0
x  4  0 OR
x40
x4
OR
x  4
Note: Instead of writing
x  16 OR x   16 ,
mathematicians use the shorthand
“plus-minus” notation of
x   16
.
x  4
Solve x 2  16  0 using the square root property.
x 2  16  0
x 2  16
x  16 OR
x   16
x4
x  4
OR
Intermediate Algebra
Test #7 Review
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Another Example: Note that the square term can be the square of a quantity.
 y  3  100  0
2
 y  3  100
2
y  3  100
OR
y  3   100
y  3  10
OR
y  3  10
y7
OR
y  13
Here is how to write the solution using “plus-minus” notation:
 y  3
2
 100
 y  3
2
  100
y  3  10
y  3  10
y  3  10 OR
y  3  10
y7
y  13
OR
When the quadratic equation is prime (i.e., it can’t be factored), like x 2  2 x  1  0 , we manipulate the
equation so that it becomes the square of a binomial plus a constant. This manipulation involves taking the first
two terms, and finding out what we have to add to them to make a perfect square trinomial, which can be
replaced with the square of a binomial.
So, for x 2  2 x  1  0 , the first two terms are identical to the first two terms of the perfect square
2
trinomial x 2  2 x  1 , which comes from the square of the binomial x + 1:  x  1  x 2  2 x  1 . So, here is
what we do:
Intermediate Algebra
Test #7 Review
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x2  2x 1  0
x2  2x  1
x2  2x  1  1  1
 x  1
2
2
 x  1
2
 2
x 1   2
x  1  2
x  1  2
OR
x  1  2
x  0.414
OR
x  2.414
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.
2

1 
i.e., Find the number b in x  bx  c and compute  b 
2 
Add that number to both sides of the equation.

1 
1 
i.e., x  bx   b   c   b 
2 
2 
Write the resulting perfect square trinomial as the square of the binomial .

1 

1 
i.e.,  x  b   c   b 
2 

2 
Use the square root property to solve the equation.
2
2
2
2
2
2
Intermediate Algebra
Test #7 Review
Page 4 of 8
Section 8.2: Solving Quadratic Equations by the Quadratic Formula
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
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.
2. If b2 – 4ac is positive but not a perfect square, then there are two solutions that are real, irrational, and
unequal.
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.
4. If b2 – 4ac is negative, then there are two solutions that are complex and unequal.
Intermediate Algebra
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Section 8.3: Solving Equations Quadratic in Form
Big Skill: You should be able to solve quadratic-like equations by making an appropriate substitution and then
solving the resulting quadratic equation.
Example:
x4  x2  6  0
Let u = x2.  u2 = (x2)2 = x4.

u2  u  6  0
 u  3 u  2   0
u  3 u  2
u  3 u  2
x 2  3 x 2  2
x 3
x   2i
Section 8.4: Graphing Quadratic Equations Using Transformations
Definition: Quadratic Function
A quadratic function (in general form) is a function of the form f  x   ax2  bx  c where a, b, and c are real
numbers and a  0. The domain consists of all real numbers. The point (0, c) is the y-intercept of the graph of
the quadratic function. The graph of a quadratic function is always a parabola.
A quadratic function (in standard form) is a function of the form f  x   a  x  h   k where a, h, and k are
2
real numbers and a  0. The point (h, k) is the vertex of the graph of the quadratic function.
The graph of every quadratic function can be obtained by transforming the graph of y = x2 with:
1. a vertical shift,
2. a horizontal shift,
3. a reflection about the x-axis,
4. and/or a vertical stretch or compression
Intermediate Algebra
Test #7 Review
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Graph of the Quadratic Function in Standard Form f  x   a  x  h   k :
2
 Locate the vertex of the graph at (h, k).
 If a > 0, sketch the parabola opening up, otherwise
sketch it opening down.
 If |a| > 0, sketch the parabola vertically stretched,
otherwise sketch it vertically compressed.
 The picture to the left shows the steps for graphing
2
y  1.5  x  3  1 :



Locate vertex at (3, -1)
Sketch parabola opening down
Sketch parabola vertically stretched.
Graph of the Quadratic Function in General Form f  x   ax2  bx  c :
 Convert the function to standard form
2
f  x   a  x  h   k by completing the square
 Graph the function f  x   a  x  h   k using
transformations.
2
 To graph y  x 2  2 x  5 :
y   x2  2x   5
y  1   x 2  2 x  1  5
y  1   x  1  5
2
y   x  1  6
2
Intermediate Algebra
Test #7 Review
Page 7 of 8
Section 8.5: Graphing Quadratic Equations Using Properties
Big Idea: There are formulas that convert between the general form and standard form of a quadratic function.
Big Skill: You should be able to use those formulas to convert between forms so that you can quickly sketch the
graph of a quadratic function
Quadratic function in general form: f  x   ax2  bx  c
Quadratic function standard form: f  x   a  x  h   k
2
h
b
2a
4ac  b 2
k
4a
The Vertex of a Parabola
Any quadratic function in general form f  x   ax2  bx  c (a  0) will have its vertex at the point whose
coordinates are:
 b 4ac  b2 
 ,
 .
4a 
 2a
Two alternative ways to state the vertex coordinates are using the discriminant:
D
 b
D  b 2  4ac    ,  
 2a 4a 
And by plugging the x-coordinate of the vertex into the function (i.e., since y = f(x) ):
 b
 b 
  2a , f   2a  



Intermediate Algebra
Test #7 Review
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The x-Intercepts of the Graph of Parabola
The x-intercepts of a graph are the x values where y = 0:
y0
f  x  0
ax 2  bx  c  0
Thus, the x-intercepts of the graph of a parabola are given by the quadratic formula. We can anticipate the
number of x-intercepts based on the discriminant:
If the discriminant D  b 2  4ac  0 , then the graph of f  x   ax2  bx  c has two different x-intercepts at
b  D
.
2a
If the discriminant D  b 2  4ac  0 , then the graph of f  x   ax2  bx  c has one x-intercept, and the vertex of
x
b
.
2a
If the discriminant D  b 2  4ac  0 , then the graph of f  x   ax2  bx  c has no x-intercepts (the graph does
not cross or touch the x-axis).
the graph will touch the x-axis at x 
To Graph a Quadratic Function Using Its Properties:
b
4ac  b 2
 Use the formulas h  
and k 
to quickly convert the general form of the quadratic equation,
2a
4a
2
f  x   ax2  bx  c , to the standard form f  x   a  x  h   k .
 Graph the standard form using translations.