Download Solving Quadratic Equations

Solving Quadratic Equations
Solving Linear Equations
• You know by now how to solve an equation such as
3𝑥 − 19 = −4
• This equation can be written in the form 𝑎𝑥 + 𝑏 = 0, and these are
called linear equations
• In fact, a linear equation is really just a first degree polynomial
equation
• Every first degree polynomial equation (or linear equation) can be
solved using only the following
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Solving Linear Equations
• To solve 3𝑥 − 19 = −4 you need
• The properties of equality: 𝑎 = 𝑏 ⟹ 𝑎 + 𝑐 = 𝑏 + 𝑐 and 𝑎 = 𝑏 ⟹ 𝑎𝑐 = 𝑏𝑐
• The inverse property of addition: 𝑎 + −𝑎 = 0
1
𝑎
• The inverse property of multiplication: 𝑎 ⋅ = 1, 𝑎 ≠ 0
• In this section you will learn methods for solving quadratic
equations or second degree polynomial equations
• A quadratic equation can be written in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0,
where 𝑎 ≠ 0
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Solving Quadratic Equations
• Recall that solving a linear equation involved using the required
properties until you arrive at the equation
1 ⋅ 𝑥 + 0 = {solution}
• You will need to arrive at the same equation in solving a quadratic
equation
• However, because quadratic equations involve an 𝑥 2 term, the
properties used to solve linear equations are not enough (though they
are still valid and may be required)
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Solving Quadratic Equations
• To demonstrate, let’s use the three properties to try and solve the
quadratic equation
5𝑥 2 − 1 = 124
• Add 1 to both sides to get
5𝑥 2 + 0 = 125
1
5
• Multiply both sides by (or divide by 5) to get
1 ⋅ 𝑥 2 + 0 = 25
• As you can see, we don’t get 1 ⋅ 𝑥 + 0 = {solution}; we must do
something about the 𝑥 2 term
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Solving Quadratic Equations
• We run into a more difficult problem trying to solve the following
quadratic equation
𝑥 2 − 7𝑥 − 12 = 0
• If you add 12 to both sides you get
𝑥 2 − 7𝑥 + 0 = 12
• But now, there is no way to combine the 𝑥 2 term and the −7𝑥 term
since they are not like terms; you’re stuck after one step!
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Solving Quadratic Equations
• You will learn several techniques for solving quadratic equations
• One technique will require that you use the definition of a square root
that we developed earlier (somewhat modified)
• Another technique will require you to factor
• A third technique uses a formula called the quadratic formula
• In general, every quadratic equation can be solved using the quadratic
formula (if the equation has a solution)
• But it is often easier to use the square root definition or factoring, so
you will eventually have to choose the most efficient method
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Solving Quadratic Equations
• It will be helpful at the beginning to classify the types of quadratic
equations by the best solution method
• The general form of a quadratic equation is 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
• The method you will use to solve an equation of this type depends on
the values of 𝑏 and 𝑐
• If 𝑏 = 0, the equation becomes 𝑎𝑥 2 + 𝑐 = 0; use the square root property (we
will define this soon)
• If 𝑐 = 0, the equations becomes 𝑎𝑥 2 + 𝑏𝑥 = 0 and this is easily solved by
factoring
• If neither 𝑏 not 𝑐 are zero, then either factor as a quadratic or use the quadratic
formula
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2
Solve Quadratic Equations: 𝑎𝑥 + 𝑐 = 0
• Recall that our definition of square root was
If 𝑦 2 = 𝑥, then 𝑦 = 𝑥
• This is the definition of the principal square root (answer is always
positive)
• We need to modify this a bit to obtain the square root property
• Notice that if 𝑦 2 = 16, then 𝑦 = 4 is a solution since 42 = 16
• But it turns out that 𝑦 = −4 is also a solution because −4 2 = −4 ⋅
− 4 = 16
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2
Solve Quadratic Equations: 𝑎𝑥 + 𝑐 = 0
• So the first thing to remember about solving a quadratic equation of
the form 𝑎𝑥 2 + 𝑐 = 0 is that there are always two solutions (unless
the solution is zero)
• Here, then, is the square root property (write it in your notes)
• If 𝑥 2 = 𝑦, then 𝑥 = 𝑦 or 𝑥 = − 𝑦
• We can see that this must be true because
𝑦 2 = 𝑦 ⋅ 𝑦 = 𝑦2 = 𝑦
• And
− 𝑦 2 = − 𝑦 ⋅ − 𝑦 = 𝑦2 = 𝑦
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2
Solve Quadratic Equations: 𝑎𝑥 + 𝑐 = 0
• The square root property is
𝑥 2 = 𝑦 ⟹ 𝑥 = 𝑦 or 𝑥 = − 𝑦
• Sometimes this is written more compactly as
𝑥2 = 𝑦 ⟹ 𝑥 = ± 𝑦
• Now let’s use the square root property to solve the equation you saw
earlier
5𝑥 2 − 1 = 125
• We were able to simplify this to 𝑥 2 = 25
• Now by the square root property, 𝑥 = ± 25 = ±5
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2
Solve Quadratic Equations: 𝑎𝑥 + 𝑐 = 0
• Before practicing a few examples, notice that, although the general
form is 𝑎𝑥 2 + 𝑐 = 0, this equation has solutions only if 𝑐 ≤ 0
• To see why, consider the equation 𝑦 = −1
• By the square root definition we must have 𝑦 2 = −1, or 𝑦 ⋅ 𝑦 = −1
• But if 𝑦 = 1, then 𝑦 2 = 𝑦 ⋅ 𝑦 = 1 ⋅ 1 = 1, not −1
• Also, if 𝑦 = −1, then 𝑦 2 = 𝑦 ⋅ 𝑦 = −1 ⋅ −1 = 1, not −1
• There is no real number such that 𝑦 2 = −1, and the same is true for
any other negative numbers
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2
Solve Quadratic Equations: 𝑎𝑥 + 𝑐 = 0
• Now, if 𝑐 > 0 in the equation 𝑥 2 + 𝑐 = 0, then we get 𝑥 2 = −𝑐 (by
adding 𝑐 to both sides of the equation) and this has no real number
answer
• So we should really think of the general form of the equation as
𝑎𝑥 2 − 𝑐 = 0
• Before we finish this chapter, we will solve the problem of −1 by
inventing some new numbers
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Guided Practice
Use the square root property to solve each equation below. If it cannot
be solved, write “no real number solution exists”
a) 2𝑥 2 − 7 = 25
b) −6𝑥 2 + 1 = −215
c) 𝑥 2 − 18 = 0
d) 2𝑥 2 + 8 = 0
e) 𝑥 2 = 98
f) 3𝑥 2 − 16 = −16
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Guided Practice
Use the square root property to solve each equation below. If it cannot
be solved, write “no real number solution exists”
a) 2𝑥 2 − 7 = 25, 𝑥 = ±4
b) −6𝑥 2 + 1 = −215, 𝑥 = ±6
c) 𝑥 2 − 18 = 0, 𝑥 = ± 18 = ±3 2
d) 2𝑥 2 + 8 = 0, no real number solution
e) 𝑥 2 = 98, 𝑥 = ± 98 = ±7 2
f) 3𝑥 2 − 16 = −16, 𝑥 = 0
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Now we turn to solving quadratic equations of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
where neither 𝑏 nor 𝑐 is zero and where the expression can be factored
• First, let’s review how to factor a quadratic expression
• To factor 8𝑥 2 − 10𝑥 + 3
• Multiply the lead coefficient by the constant to get 24
• Find the factors of 24 that add to −10: since −6 ⋅ −4 = 24 and −6 + −4 = −10,
we use these
• Rewrite the expression as 8𝑥 2 − 6𝑥 − 4𝑥 + 3 and group as 8𝑥 2 − 6𝑥 + −4𝑥 + 3
• Factor the GCF from each group: 2𝑥 4𝑥 − 3 − 1 ⋅ 4𝑥 − 3 = (4𝑥 − 3)(2𝑥 − 1)
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• To solve the quadratic equation 8𝑥 2 − 10𝑥 + 3 = 0 (note that it must
equal zero!), we use a property of numbers involving zero
• Product Property of Zero: If 𝑎𝑏 = 0, then 𝑎 = 0 or 𝑏 = 0
• That is, if you multiply two numbers and the result is zero, then one or
the other (or both) of the numbers must be zero
• We can use this to solve by factoring because factoring an expression
makes the expression a multiplication
• This is why the equation must be equal to zero!!!
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Let’s use this property to solve 8𝑥 2 − 10𝑥 + 3 = 0
• You already know that 8𝑥 2 − 10𝑥 + 3 = 4𝑥 − 3 2𝑥 − 1
• So we rewrite the equation as 4𝑥 − 3 2𝑥 − 1 = 0
• Using the zero property 𝑎𝑏 = 0 ⟹ 𝑎 = 0 or 𝑏 = 0, the expression
4𝑥 − 3 is like the 𝑎 and the expression 2𝑥 − 1 is like the 𝑏
• At this point, we can split the equation into two equations
• Either 4𝑥 − 3 = 0 or 2𝑥 − 1 = 0 (note that these are both linear equations)
• Finally, solve each linear equation to get 𝑥 =
3
4
or 𝑥 =
1
2
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Now check the results
3
8
4
2
3
− 10
+3=
4
9
30
8
−
+3=
16
4
9 30
18 30
−12
−
+3=
−
+3=
+ 3 = −3 + 3 = 0
2 4
4
4
4
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Now check the results
1
8
2
2
1
− 10
+3=
2
1
8
−5+3=
4
2 − 5 + 3 = −3 + 3 = 0
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Example: solve 𝑥 2 − 2𝑥 = 48
• The equation must equal zero; subtract 48 from both sides of the equation
𝑥 2 − 2𝑥 − 48 = 0
• Factor the left side of the equation
𝑥+6 𝑥−8 =0
• Create two linear equations by setting each factor equal to zero (the product
property of zero)
𝑥 + 6 = 0 or 𝑥 − 8 = 0
• Solve each linear equation
𝑥 = −6 or 𝑥 = 8
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Example: solve 𝑥 2 = 8𝑥 − 16
• The equation must equal zero; subtract 8𝑥 and add 16
𝑥 2 − 8𝑥 + 16 = 0
• Factor the left side of the equation
𝑥−4 𝑥−4 =0
• Create two linear equations by setting each factor equal to zero (the product
property of zero)
𝑥 − 4 = 0 or 𝑥 − 4 = 0
• Solve each linear equation (note that this equation has only one solution)
𝑥=4
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• We can also use the square root property to solve this equation (after it
has been factored)
• Since (𝑥 − 4)(𝑥 − 4) is the same as 𝑥 − 4 2 , then the equation
becomes
𝑥−4 2 =0
• By the square root property,
𝑥 − 4 = ±0 = 0
• Therefore, 𝑥 = 4 is the only solution
• We will use the square root property in this way when you learn one
more method for solving quadratic equations
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Example: solve 𝑥 2 − 2𝑥 − 2 = 0
•
•
•
•
•
•
This is already equal to zero, so no change is needed in the equation
There are no integers that multiply to give −2 and also add to give −2
This does not mean that this has no solution!
It means that we cannot solve this by factoring
The two solutions are 𝑥 = 1 + 2 and 𝑥 = 1 − 2
You will learn how to solve these by other methods (and this type of problem
will not appear in your practice)
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Example: solve 2𝑥 2 + 7𝑥 + 3 = 0
• This is already equal to zero, so no change is needed in the equation
• Multiply the lead coefficient by the constant to get 6: choose 6 and 1 since 6 ⋅
1 = 6 and 6 + 1 = 7
• Rewrite the equation as 2𝑥 2 + 6𝑥 + 𝑥 + 3 = 0
• Factor by grouping
2𝑥 2 + 6𝑥 + 𝑥 + 3 = 2𝑥 𝑥 + 3 + 1 𝑥 + 3 = (2𝑥 + 1)(𝑥 + 3)
• The equation is now 2𝑥 + 1 𝑥 + 3 = 0
• Set each factor equal to zero and solve the two linear equations
2𝑥 + 1 = 0 or 𝑥 + 3 = 0
1
• The solutions are 𝑥 = − or 𝑥 = −3
2
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Example: solve −10𝑐 2 = 11𝑐 − 6
• This is not equal to zero; it is easier to factor if the lead coefficient is not
negative, so add 10𝑐 2 to both sides
0 = 10𝑐 2 + 11𝑐 − 6 or 10𝑐 2 + 11𝑐 − 6 = 0
• Factor the expression: 10 ⋅ −6 = −60; choose 15 and −4 since 15 ⋅ −4 =
− 60 and 15 + −4 = 11
• Factor by grouping
10𝑐 2 + 15𝑐 − 4𝑐 − 6 = 0
10𝑐 2 + 15𝑐 + −4𝑐 − 6 = 0
5𝑐 2𝑐 + 3 − 2 2𝑐 + 3 = 0
2𝑐 + 3 5𝑐 − 2 = 0
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2
Solving 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 by Factoring
• Example: solve −10𝑐 2 = 11𝑐 − 6
• Set each factor equal to zero and solve the linear equations
5𝑥 − 2 = 0 or 2𝑥 + 3 = 0
2
3
𝑥 = or 𝑥 = −
5
2
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2
Solving 𝑎𝑥 + 𝑏𝑥 = 0
• Note that a quadratic equation of the form 𝑎𝑥 2 + 𝑏𝑥 = 0 is easy to
solve because we will always be able to factor an 𝑥
• Example: solve 4𝑥 2 − 8𝑥 = 0
• We can find the GCF and rewrite the equation as
4𝑥 𝑥 − 2 = 0
• Now we again apply the zero product property with 𝑎 being 4𝑥 and 𝑏
as 𝑥 − 2
4𝑥 = 0 or 𝑥 − 2 = 0
𝑥 = 0 or 𝑥 = 2
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Guided Practice
Solve the following quadratic equations by factoring.
a) 6𝑥 2 = 5𝑥
b) 𝑥 2 − 8𝑥 + 15 = 0
c) 10𝑥 − 3 = −8𝑥 2
d) 7𝑥 2 + 9𝑥 = −2
e) 𝑥 2 − 12𝑥 + 36 = 0
f) 8𝑥 2 − 16𝑥 = 0
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Guided Practice
Solve the following quadratic equations by factoring.
a)
6𝑥 2
= 5𝑥, 𝑥 = 0 or 𝑥 =
5
6
b) 𝑥 2 − 8𝑥 + 15 = 0, 𝑥 = 3 or 𝑥 = 5
c) 10𝑥 − 3 = −8𝑥
−3
1
,𝑥=
or 𝑥 =
2
4
−2
𝑥=
or 𝑥 = −1
7
2
d) 7𝑥 2 + 9𝑥 = −2,
e) 𝑥 2 − 12𝑥 + 36 = 0, 𝑥 = 6
f) 8𝑥 2 − 16𝑥 = 0, 𝑥 = 0 or 𝑥 = 2
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