```Martin-Gay, Developmental Mathematics
1
Solving Quadratic Equation by Square Root Property



We previously have used factoring to solve quadratic equations.
This chapter will introduce additional methods for solving
Square Root Property

If b is a real number and a2 = b, then
Example
♦ Solve x2 = 49
x   49  7
♦ Solve 2x2 = 4
x2 = 2
x 2
♦ Solve (y – 3)2 = 4
y  3   4  2
y=32
y = 1 or 5
a b
♦ Solve x2 + 4 = 0
x2 = 4
There is no real solution
because the square root of 4
is not a real number.
Example
Solve (x + 2)2 = 25
x  2   25  5
x = 2 ± 5
x = 2 + 5 or x = 2 – 5
x = 3 or x = 7
Solve (3x – 17)2 = 28
3x – 17 =  28   2 7
3x  17  2 7
17  2 7
x
3
Solving quadratic Equation by Completing the Square
In all four of the previous examples, the constant in the square
on the right side, is half the coefficient of the x term on the left.
Also, the constant on the left is the square of the constant on the
right.
So, to find the constant term of a perfect square trinomial, we
need to take the square of half the coefficient of the x term in the
trinomial (as long as the coefficient of the x2 term is 1, as in our
previous examples).
Example
What constant term should be added to the following expressions
to create a perfect square trinomial?
x2 – 10x
52
x2 – 7x
= 25
x2 + 16x
2
7
49
4
2
Example
We now look at a method for solving quadratics that involves a
technique called completing the square.
It involves creating a trinomial that is a perfect square, setting
the factored trinomial equal to a constant, then using the square
root property from the previous section.
Solving a Quadratic Equation by Completing a Square
1) If the coefficient of x2 is NOT 1, divide both sides of the
equation by the coefficient.
2) Isolate all variable terms on one side of the equation.
3) Complete the square (half the coefficient of the x term squared,
added to both sides of the equation).
4) Factor the resulting trinomial.
5) Use the square root property.
Example
Solve by completing the square.
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y2 + 6y = 8
y + 3 = ± 1= ± 1
y = 3 ± 1
y = 4 or 2
Solve by completing the square.
y2 + y = 7
y2 + y + ¼ = 7 + ¼
29
(y + ½)2 =
4
1
29
29
y 

2
4
2
1
29  1  29
y 

2
2
2
y2 + y – 7 = 0
♦ Another technique for solving quadratic equations is to use the
♦ The formula is derived from completing the square of a general
♦ A quadratic equation written in standard form, ax2 + bx + c = 0,
has the solutions.
Example
 b  b 2  4ac
x
2a
♦ Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1
9  (9) 2  4(11)(1) 9  81 44 9  125 9  5 5


n

22
22
22
2(11)
The Discriminant
called the discriminant.
♦The discriminant will take on a value that is positive, 0, or negative.
♦The value of the discriminant indicates two distinct real solutions,
one real solution, or no real solutions, respectively.
Example
Use the discriminant to determine the number and type of
solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions.
1.
2.
3.
4.
If the equation is in the form (ax + b)2 = c, use the square
root property to solve.
If not solved in step 1, write the equation in standard form.
Try to solve by factoring.
If you haven’t solved it yet, use the quadratic formula.
Example
♦ Solve 12x = 4x2 + 4.
0 = 4x2 – 12x + 4
0 = 4(x2 – 3x + 1)
Let a = 1, b = -3, c = 1
♦ Solve the following
5 2
1
m m 0
8
2
5m2  8m  4  0
3  (3) 2  4(1)(1)
x
2(1)
(5m  2)(m  2)  0
3 9 4
3 5


2
2
5m  2  0 or m  2  0
2
m  or m  2
5
♦The graph of a quadratic equation is a parabola.
♦The highest point or lowest point on the parabola is the vertex.
Example
Graph
x
y=
2x2
y
– 4.
(–2, 4)
y
2
4
1
–2
0
–4
–1
–2
–2
4
(2, 4)
x
(–1, – 2)
(1, –2)
(0, –4)
Intercepts of the Parabola
Although we can simply plot points, it is helpful to know
some information about the parabola we will be graphing prior to
finding individual points.
To find x-intercepts of the parabola, let y = 0 and solve for x.
To find y-intercepts of the parabola, let x = 0 and solve for y.
Characteristics of the Parabola
♦ If the quadratic equation is written in standard form,
y = ax2 + bx + c,
1) the parabola opens up when a > 0 and opens down when a < 0.
2) the x-coordinate of the vertex is 
b
.
2a
To find the corresponding y-coordinate, you substitute the x-coordinate
into the equation and evaluate for y.
Example
Graph
Since a = –2 and b = 4, the
graph opens down and the xcoordinate of the vertex is
4

1
2(2)
x
y
3
–1
2
5
1
7
0
5
–1
–1
y = –2x2 + 4x + 5.
y
(0, 5)
(–1, –1)
(1, 7)
(2, 5)
(3, –1)
x
Domain and Range
Recall that a set of ordered pairs is also called a relation.
The domain is the set of x-coordinates of the ordered pairs.
The range is the set of y-coordinates of the ordered pairs.
Example
Find the domain and range of the relation
{(4,9), (–4,9), (2,3), (10, –5)}
♦ Domain is the set of all x-values, {4, –4, 2, 10}
♦ Range is the set of all y-values, {9, 3, –5}
Example
Find the domain and range of the function graphed to the
right. Use interval notation.
y
Domain
x
Domain is [–3, 4]
Range is [–4, 2]
Range
Find the domain and range of the function graphed to
the right. Use interval notation.
y
Range
x
Domain is (– , )
Range is [– 2, )
Domain
Graphing Piecewise-Defined Functions
Example
3x  2 if x  0

Graph f (x)  
.
 x  3 if x  0
Graph each “piece” separately.
Values  0.
x
f (x) = 3x – 1
x
f (x) = x + 3
0
– 1(closed circle)
1
4
2
5
3
6
–1 – 4
–2 – 7
Values > 0.
Continued.
Example continued
y
x
f (x) = 3x – 1
0
– 1(closed circle)
–1 – 4
–2 – 7
(3, 6)
Open circle
(0, 3)
(0, –1)
x
f (x) = x + 3
1
4
2
5
3
6
(–1, 4)
(–2, 7)
x
The End
Martin-Gay, Developmental Mathematics
18
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