Download The Quadratic Equation

Quadratic Equations Module
The objectives of this module are to:
1) Recognize quadratic equations &
2) Solve quadratic equations
Purpose
Since you have previously learned how to solve
a quadratic equation in one of your mathematics
courses, the purpose of this module is simply to
refresh your memory. When solving problems in
the various fields of science, frequently the
quadratic equation is encountered. A ready
ability to solve quadratic equations will make the
solution of these problems much easier.
Recognizing the Quadratic Equation
The general form of the quadratic equation is:
ax2 + bx + c = 0 (Equation 1)
where x is the variable (unknown) and a, b, and c
are known constants.
Recognizing the Quadratic Equation
Equation 2 shows that the general form of the
quadratic equation (Equation 1) actually contains
terms with the variable (x) raised to the second,
first, and zero powers, since x1 = x and x0 = 1.
ax2 + bx1 + cx0 = 0 (Equation 2)
Recognizing the Quadratic Equation
When solving a problem involving a quadratic
equation, the first step is to recognize the
presence of the quadratic equation and then to
put it in the general form.
Equation 2 showed that a quadratic equation
will exist only when a single variable raised to
the second, first, and zero power is present.
Other Not-Quadratic Equations
If only a single variable is present raised to the
first power, a linear equation (ax + b = 0) exists
and is solved in a straightforward manner.
If only a single variable raised to the second power
is present, the equation is solved by taking the
square root.
ax2 + b = 0
Transposed: ax2 = -b
Transposed: x2 = -b/a
𝒙 = ± −𝒃 𝒂
Other Not-Quadratic Equations
If two variables are present (raised to the first
power), a second equation must be present so the
equations can be solved simultaneously.
ax + by + c = 0
dx + ey + f = 0
(x and y are variables)
& (a, b, c, d, e, and f are constants)
Recognizing the Quadratic Equation
Again, a quadratic equation will exist only when a
single variable raised to the second, first, and zero
power is present.
Let's look at several examples and determine
whether the equation is a quadratic by attempting
to put it in the general form.
Example 1
• Equation: 4x2 + 3x = 7x + 10
• Transposed: 4x2 + 3x - 7x - 10 = 0
• Combined: 4x2 - 4x - 10 = 0
Yes, this is a quadratic equation because it
follows the form of Equation 1.
Example 2
• Equation: 8 = (2a-7)/(-3a)
[a is the variable]
• Multiplied by -3a: -24a = 2a – 7
• Transposed: 0 = 24a + 2a – 7
• Combined: 26a - 7 = 0
This equation is not a quadratic; it is a linear
equation
Example 3
• Equation: (c - 2) (c + 3) = 2 [c is the variable]
• Expanded: c2 – 2c + 3c – 6 = 2
• Transposed: c2 - 2c + 3c - 6 - 2 = 0
• Combined: c2 + c - 8 = 0
Yes, this is a quadratic equation.
Example 4
• Equation: 1 = z/(z + 2)2
[z is the variable]
• Multiplied by (z + 2)2: (z + 2)2 = z
• Expanded: z2 + 4z + 4 = z
• Transposed: z2 + 4z - z + 4 = 0
• Combined: z2 + 3z + 4 = 0
Yes, this is a quadratic equation.
Practice Set 1
Now using a piece of scratch paper, determine
whether or not the following equations are
quadratic equations. The answers are at the end of
the module.
1) Equation: B2 + 2 = 10B (B + 1)
2) Equation: Q = (5Q - 3)/Q
3) Equation: T = 4/(3 - T)2
4) Equation: 10-3 = A2/(10-1 - A)
Solving the Quadratic Equation
• The solution of the quadratic equation in the
general form ax2 + bx + c = 0 is:
−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄
𝒙=
𝟐𝒂
The result of this solution will yield two (+/-)
values or roots.
In the broadest terms, these roots can be real or
imaginary (as when (b2 - 4ac) is a negative
number).
Solving the Quadratic Equation
• The solution of the quadratic equation in the general
form ax2 + bx + c = 0 is:
−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄
𝒙=
𝟐𝒂
 However, in the real world—which most problems in
science courses deal with—the correct value is not
imaginary and most often positive
 Example: if the variable (x) is the mass of a
substance, a negative value would be meaningless
since there cannot be negative mass
Solving the Quadratic Equation
• The solution of the quadratic equation in the
general form ax2 + bx + c = 0 is:
−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄
𝒙=
𝟐𝒂
Care must be taken in selecting the correct root
to the quadratic equation so the answer to the
problem is meaningful.
Example 5
• Now let's work through several examples solving
for the roots of the quadratic equation.
• Equation: x2 - 5x + 4 = 0
a =1
b = -5
c=4
−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄
𝒙=
𝟐𝒂
Example 5 cont’d
−(−𝟓) ± (−𝟓)𝟐 −𝟒(𝟏)(𝟒) 𝟓 ± 𝟐𝟓 − 𝟏𝟔
𝒙=
=
𝟐(𝟏)
𝟐
𝟓± 𝟗 𝟓±𝟑
=
=
𝟐
𝟐
if x = (5 + 3)/2 then x = 8/2 = 4
if x = (5 - 3)/2 then x = 2/2 = 1
Example 6
Equation: x2 + 10x + 2 = 0
−𝟏𝟎 ± 𝟏𝟎𝟐 − 𝟒(𝟏)(𝟐) −𝟏𝟎 ± 𝟏𝟎𝟎 − 𝟖
𝒙=
=
𝟐(𝟏)
𝟐
−𝟏𝟎 ± 𝟗𝟐 −𝟏𝟎 ± 𝟗. 𝟓𝟗
=
=
𝟐
𝟐
if x = (-10 + 9.59)/2 then x = -0.41/2 = -0.21
if x = (-10 - 9.59)/2 then x = -19.59/2 = -9.8
Practice Set 2
On a piece of scratch paper solve these equations
using the quadratic formula. The answers are at
the end of the module.
1) Equation: 4x2 + 3x - 1 = 0
2) Equation: a2 + (2.0 x 10-4)a - (1.5 x 10-8) = 0
3) Equation: 4x2 - 0.2125x + 0.00125 = 0
ANSWERS
Answers to Practice Set 1
1) Equation: B2 + 2 = 10B (B + 1)
Expand: B2 + 2 = 10B2 + 10B
Transpose: B2 - 10B2 - 10B + 2 = 0
Combine: -9B2 - 10B + 2 = 0
Yes, this is a quadratic equation.
ANSWERS
Answers to Practice Set 1
2) Equation: Q = (5Q - 3)/Q
Transpose: Q2 = 5Q – 3
Transpose: Q2 - 5Q + 3 = 0
Yes, this is a quadratic equation.
ANSWERS
Answers to Practice Set 1
3) Equation: T = 4/(3 - T)2
Transpose: T(3 - T)2 = 4
Expand: T(9 - 6T + T2) = 4
Expand: 9T - 6T2 + T3 = 4
Transpose: T3 - 6T2 + 9T - 4 = 0
This is not a quadratic equation
ANSWERS
Answers to Practice Set 1
4) Equation: 10-3 = A2/(10-1 - A)
Transpose: 10-4 - 10-3A = A2
Transpose: A2 + 10-3A - 10-4 = 0
Yes, this is a quadratic equation.
ANSWERS
Answers to Practice Set 2
1) Equation: 4x2 + 3x - 1 = 0
𝒙=
−𝟑± (𝟑)𝟐 −𝟒(𝟒)(−𝟏)
𝟐(𝟒)
−𝟑± 𝟐𝟓
𝟖
=
=
−𝟑±𝟓
𝟖
x = 2/8 = 0.25
AND
x = -8/8 = -1
−𝟑± 𝟗+𝟏𝟔
𝟖
=
ANSWERS
Answers to Practice Set 2
2) Equation: a2 + (2.0 x 10-4)a - (1.5 x 10-8) = 0
(−𝟐 × 𝟏𝟎−𝟒 ) ± (𝟐. 𝟎 × 𝟏𝟎−𝟒 )𝟐 −𝟒(𝟏)(−𝟏. 𝟓 × 𝟏𝟎−𝟖 )
𝒂=
𝟐(𝟏)
−𝟐 × 𝟏𝟎−𝟒 ± 𝟒. 𝟎 × 𝟏𝟎−𝟖 + 𝟔. 𝟎 × 𝟏𝟎−𝟖
𝒂=
𝟐
−𝟐 × 𝟏𝟎−𝟒 ± 𝟏𝟎. 𝟎 × 𝟏𝟎−𝟖 −𝟐 × 𝟏𝟎−𝟒 ± (𝟑. 𝟐 × 𝟏𝟎−𝟒 )
𝒂=
=
𝟐
𝟐
𝒂=
𝟏.𝟐×𝟏𝟎−𝟒
𝟐
= 𝟔. 𝟎 ×
𝟏𝟎−𝟓
AND 𝒂 =
−𝟓.𝟐×𝟏𝟎−𝟒
𝟐
= −𝟐. 𝟔 × 𝟏𝟎−𝟒
ANSWERS
Answers to Practice Set 2
3) Equation: 4x2 - 0.2125x + 0.00125 = 0
−(−𝟎. 𝟐𝟏𝟐𝟓) ± (−𝟎. 𝟐𝟏𝟐𝟓)𝟐 −𝟒(𝟒)(𝟎. 𝟎𝟎𝟏𝟐𝟓)
𝒙=
𝟐(𝟒)
𝟎. 𝟐𝟏𝟐𝟓 ± 𝟎. 𝟎𝟒𝟓𝟐 − 𝟎. 𝟎𝟐𝟎𝟎
=
𝟖
𝟎. 𝟐𝟏𝟐𝟓 ± 𝟎. 𝟎𝟐𝟓𝟐 𝟎. 𝟐𝟏𝟐𝟓 ± 𝟎. 𝟏𝟓𝟖𝟕
=
=
𝟖
𝟖
𝒙=
𝟎.𝟑𝟕𝟏𝟐
𝟖
= 𝟎. 𝟎𝟒𝟔𝟎 AND 𝒙 =
𝟎.𝟎𝟓𝟑𝟖
𝟖
= 𝟎. 𝟎𝟎𝟔𝟕𝟐𝟓
Done!
If you have any questions or are having difficulty
solving these problems, ask the assistant in the
Science Learning Center for help. If you feel
competent in solving the quadratic equation, ask
the assistant for the post-test.