# Section 11.1 - The Square Root Property and Completing the Square ```Math 127 - Section 11.1 - Page 1
Section 11.1 - The Square Root Property and
Completing the Square
I.
Goal
2
A.
We want to get every quadratic equation (Ax + Bx + C = 0) to be of the form:
Perfect Square = A Constant
1.
The perfect square will be some kind of variable expression.
2.
The constant clearly will have no variables.
B. Once we have the equation in this form, we will take the square root of each side.
However, since we are taking the square root, we need to remember that every real
number has both a positive and a negative square root. So we need to make sure
that when we take the square root (when solving) to write “+” in front of the square
root of the constant.
II.
Examples – Solve each of the following using the square root property.
2
A.
(4y – 6) = 48
We notice that the left side of the equation is a perfect square and the right side is a
constant. Thus we can take the square root of both sides to get:
( 4y − 6 )
2
= &plusmn; 48
The left side simplifies nicely to 4y – 6, but the right side is not a perfect square, but
we can simplify it but dividing it up into the perfect square part and the non-square
part to get:
4y − 6 = &plusmn; 16 3
OR
4y − 6 = &plusmn;4 3
Now we need to solve for y. First add 6 to both sides to get:
4y = 6 &plusmn; 4 3
Now divide both sides by 4 to get:
6&plusmn;4 3
y=
4
Factor out the GCF in the numerator to get:
(
2 3&plusmn;2 3
y=
)
4
We now can reduce to get:
3&plusmn;2 3
y=
2
B.
(2a + 6) = −64
We have a perfect square equal to a constant, so we can use the Square Root
Property:
( 2a + 6 )
2
2
= &plusmn; −64
OR
2a + 6 = + 8i
Now, we subtract 6 from both sides, but we write it a little bit weird:
2a = -6 + 8i
Now divide 2 from both sides to get:
−6 8i
a=
&plusmn;
2
2
a = −3 + 4i
Math 127 - Section 11.1 - Page 2
C.
Now you try one: (y – 5) = −27
y = 5 &plusmn; 3i 3
2
2(x − 3) – 50 = 0
We first have to get the constant on the right hand side of the equation by itself by
2
2(x − 3) = 50
Now we divide both sides by 2 to get a perfect square equal to a constant:
2
(x − 3) = 25
Now we can use the Square Root Property:
2
D.
( x − 3)
2
= &plusmn; 25
OR
x−3=+5
Adding 3 to both sides, but we have to remember that it is not got a &quot;+&quot; on it:
x=3+5
This means:
x = 3 + 5 OR x = 3 − 5
x = −2, 8
E.
Now you try one:
x=+4
F.
2
x = 16
Of the one hundred largest economies in the world, 53 are multinational
corporations. In 1970, there were approximately 7,000 multinational corporations.
By 2001, more than 65,000 corporations enveloped the world. The graph (right
hand column, Page 749) shows this rapid growth from 1970 through 2001, including
the starting dates of some notable corporations.
The data shown can be modeled by the function
f(x) = 62.2x2 + 7,000
where f(x) represents the number of multinational corporations in the world x years
after 1970. Use this function and the Square Root Property to solve exercises 109 110.
In which year were there 46,000 multinational corporations? (#110)
We first set f(x) = 46,000 and solve the equation:
2
46,000 = 62.2x + 7,000 Subtract 7,000 from both sides.
2
39,000 = 62.2x
Divide 62.2 from both sides.
2
627.0096463 = x
Take the square root of both sides.
&plusmn; 627.0096463 =
25.04016067 ≈ x
x2
We can ignore the &quot;-&quot; in the &quot;+&quot; (why?).
Round to the nearest year and add to 1970.
Math 127 - Section 11.1 - Page 3
G.
A supporting wire is to be attached to the top of a 70-foot antenna. If the wire must
be anchored 70 feet from the base of the antenna, what length of wire is required?
(Page 750, #118)
I will begin by drawing a diagram, letting c = the distance from the ground to the top
of the antenna.
Antenna
c feet
70 feet
70 feet
As we can see, this is a right triangle, so we will use the Pythagorean Theorem:
a 2 + b 2 = c2
2
2
2
(70) + (70) = c
2
4900 + 4900 = c
2
9,800 = c
9,800 = c 2
c = 98.99494937 ≈ 99
H..
Now you try Page 750, #113
.
A rectangular park is 6 miles long and 3 miles wide. How long is a pedestrian route
that runs diagonally across the park?
III.
Definitions
A.
1. In any expression, the term that has an exponent of 2 on the variable is the
2.
Examples
2
2
a.
In 5x – 3x + 2, 5x is the quadratic term.
2
2
b.
In 0.75x + 8.15x – 22.37, 0.75x is the quadratic term.
B.
Linear Term
1. In any expression, the term that has an exponent of 1 on the variable is the
linear term.
2.
Examples
2
a.
In 5x – 3x + 2, -3x is the linear term.
2
b.
In 0.75x + 8.15x – 22.37, 8.15x is the linear term.
Math 127 - Section 11.1 - Page 4
IV.
C.
Constant Term
1.
In any expression, the term that has no variable is the constant term.
2.
Examples
2
a.
In 5x – 3x + 2, 2 is the constant term.
2
b.
In 0.75x + 8.15x – 22.37, −22.37 is the constant term.
D.
A quadratic equation is any equation that is of degree 2 (i.e. – the biggest
exponent is 2) in one variable. In standard form it will look like:
2
Ax + Bx + C = 0
Where A is not 0.
Solving by Completing the Square
A.
Procedure
1.
Get the quadratic and linear terms on one side of the equation, the constant
term on the other.
2.
Make the coefficient of the quadratic term 1 by dividing everything by its
coefficient.
3.
Take half the coefficient of the linear term; square it; add it to both sides.
4.
You have now completed the square and can solve the equation using the
Square Root Property.
B.
Examples – Solve each of the following by completing the square.
2
1.
x – 2x – 8 = 0
We have to first move the –8 by adding 8 to both sides to get:
2
x – 2x = 8
The coefficient of the quadratic term is already 1, so we can take half the
coefficient of the linear term (1); square it (1); and add it to both sides to get:
2
x – 2x + 1 = 8 + 1
OR
2
(x – 1) = 9
Taking the square root of both sides, we get:
x – 1 = +3
Adding 1 to both sides, we get:
x=1+3
So we have:
x = 1 + 3 OR x = 1 – 3
x = −2, 4
2.
2
x + 10x + 1 = 0
First we subtract 1 from both sides to get:
2
x + 10x = −1
Now we take half of the linear coefficient (10); square it (25); add it to both
sides to get:
2
x + 10x + 25 = −1 + 25
OR
2
(x + 5) = 24
Taking the square root of both sides, we get:
x + 5 = &plusmn; 24
We have to simplify the square root, so we get:
x + 5 = &plusmn;2 6
Now we can subtract 5 from both sides to get:
x = 5 − &plusmn;2 6
Math 127 - Section 11.1 - Page 5
3.
Now you try one:
a = 1, 7
2
a – 8a + 7 = 0
2
4.
4x + 4x – 3 = 0
First, add 3 to both sides to get:
2
4x + 4x = 3
2
Now, since the coefficient of x is not 1, we need to divide both sides by 4 to
get:
x2 + x =
3
4
Now take half of the coefficient of the linear term (1); square it
1
  ; add it to
4
both sides to get:
1 3 1
x2 + x + = +
4 4 4
OR
2
1

x+  =1
2


Taking the square root of both sides, we get:
1
x + = &plusmn;1
2
1
Subtracting
from both sides, we get:
2
1
x = − &plusmn;1
2
OR
1
1
x = − + 1 OR x = − − 1
2
2
3 1
x=− ,
2 2
5.
2
3x + 12x + 6 = 0
First subtract 6 from both sides to get:
2
3x + 12x = −6
Now divide both sides by 3 to get:
2
x + 4x = −2
Now take half of the linear coefficient (4); square it (4); add it to both sides to
get:
2
x + 4x + 4 = −2 + 4
OR
2
(x + 2) = 2
Now take the square root of each side to get:
x+2=&plusmn; 2
Subtract 2 from each side to get: