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Page 1 of 7
Activity 2.2.6a Deriving the Quadratic Formula
Directions: In Column 1, you are given precise instructions on the steps you need to take in
order to “complete the square” to solve the quadratic equation below. In Column 2, show the
manipulation you are making to the quadratic equation using numbers. In Column 3, show the
exact same manipulation you perform in Column 2, but to a quadratic equation in standard form
with the variables a, b, and c.
Solve the quadratic equation 5x 2 + 6x - 2 = 0 .
Column 1: Steps to
Complete the Square
Step 1: Isolate the x terms
by using the Addition
Property of Equality.
Column 2: Show with Numbers
for 5x 2 + 6x - 2 = 0
5x 2 + 6x - 2 + 2 = 0 + 2
Step 2: Change the
coefficient of the quadratic
term to 1 by dividing both
sides of the equation by the
coefficient of the quadratic
term.
5x 2 6
2
+ x=
5 5
5
6
2
x2 + x =
5
5
Step 3: Divide the
coefficient of the linear term
in half and square it. Add
the result to both sides of the
equation using the Addition
Property of Equality. This
creates a perfect square
trinomial, which can be
factored into the square of a
binomial.
6 1 6
= × =
2 5 2 10
2
æ6ö
36
ç ÷ =
è 10 ø 100
6
36 2 36
x2 + x +
= +
5
100 5 100
Activity 2.2.6a
Column 3: Show with
Variables for ax 2 + bx + c = 0
5x 2 + 6x = 2
6
5
Connecticut Core Algebra 2 Curriculum Version 3.0
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Date:
Step 4: Factor the perfect
square trinomial you just
created into the square of a
binomial.
æ
6 ö 2 36
çx + ÷ = +
è 10 ø 5 100
Find a common denominator
to add together the rational
expressions.
æ
6ö
40 36
+
çx + ÷ =
è 10 ø 100 100
Take the square root of both
sides of the equation and
simplify.
æ
6ö
76
çx + ÷ = ±
è 10 ø
100
6
76
x+ =±
10
10
Solve for x and simplify the
expression.
Page 2 of 7
2
2
æ
6ö
76
çx + ÷ =
è 10 ø 100
2
2
6
76
±
10 10
-6 + 76 -6 - 76
x=
,
10
10
x=-
You have just derived the quadratic formula!!!
Activity 2.2.6a
Connecticut Core Algebra 2 Curriculum Version 3.0
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Page 3 of 7
NOTE: A quadratic equation in standard form is expressed as ax 2 + bx + c = 0 . For the given
quadratic equation a = 5 , b = 6 and c = -2 . Substitute these values into the quadratic formula
and simplify your result below.
x=
-(6) ± (6)2 - 4(5)(-2)
2(5)
-6 ± 36 + 40
10
-6 ± 76
x=
10
-6 + 76 -6 - 76
x=
,
10
10
x=
Does your result match the result in the last box of column two?
Activity 2.2.6a
Connecticut Core Algebra 2 Curriculum Version 3.0
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Page 4 of 7
Applying the Quadratic Formula
-b ± b2 - 4ac
x=
2a
Directions: Now that you have derived the quadratic formula, use it to solve the following
quadratic equations below. First, identify the coefficients a, b, and c in the spaces given. Then
substitute those values into the quadratic formula and simplify the expression.
Example 1
8x 2 + 22x - 21 = 0
a = 8, b = 22, c = -21
x=
-(22) ± 22 - 4(8)(-21)
2(8)
-11x 2 + 6x + 3 = 0
a = -11, b = 6, c = 3
2
-22 ± 484 + 672
16
-22 ± 1156
x=
16
-22 ± 34
x=
16
-22 + 34 -22 - 34
x=
,
16
16
12 -56
x= ,
16 16
3 -7
x= ,
4 2
x=
1.
Example 2
x 2 +10x + 25 = 0
a = ______ b = ______ c = ______
Activity 2.2.6a
x=
-(6) ± (6)2 - 4(-11)(3)
2(-11)
-6 ± 36 +132
-22
-6 ± 168
x=
-22
-6 ± 2 42
x=
-22
-6 + 2 42 -6 - 2 42
x=
,
-22
-22
3 - 42 3+ 42
x=
,
11
11
x=
2.
x 2 - 4x + 4 = 0
a = ______ b = ______ c = ______
Connecticut Core Algebra 2 Curriculum Version 3.0
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3.
Date:
x2 - x - 6 = 0
a = ______ b = ______ c = ______
5.
6x 2 +11x + 3 = 0
a = ______ b = ______ c = ______
7.
2x 2 + 8x + 3 = 0
a = ______ b = ______ c = ______
Activity 2.2.6a
Page 5 of 7
4.
x 2 - 2x - 35 = 0
a = ______ b = ______ c = ______
6.
5x 2 - 21x + 4 = 0
a = ______ b = ______ c = ______
8.
3x 2 - 6x -10 = 0
a = ______ b = ______ c = ______
Connecticut Core Algebra 2 Curriculum Version 3.0
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Page 6 of 7
Recall that the square root of a negative number in not a Real Number. Therefore, if your
quadratic formula results with a negative number inside the square root symbol, then the answer
is “no Real Number solution”. Some of the following equations have a real number solution, and
some do not:
9. 2 x 2  x  3  0
10.
a = ______ b = ______ c = ______
11. x 2  9  0
a = ______ b = ______ c = ______
Activity 2.2.6a
2x2  x  3  0
a = ______ b = ______ c = ______
12.
x2  9  0
a = ______ b = ______ c = ______
Connecticut Core Algebra 2 Curriculum Version 3.0
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Page 7 of 7
13. 5 x 2  8 x  5  0
a = ______ b = ______ c = ______
Activity 2.2.6a
Connecticut Core Algebra 2 Curriculum Version 3.0