quadratic equations and functions - anna kuczynska...350 quadratic equations and functions in this...
TRANSCRIPT
350
Quadratic Equations and Functions
In this chapter, we discuss various ways of solving quadratic equations, πππ₯π₯2 + πππ₯π₯ +ππ = 0, including equations quadratic in form, such as π₯π₯β2 + π₯π₯β1 β 20 = 0, and solving formulas for a variable that appears in the first and second power, such as ππ in ππ2 β 3ππ = 2ππ. Frequently used strategies of solving quadratic equations include the completing the square procedure and its generalization in the form of the quadratic formula. Completing the square allows for rewriting quadratic functions in vertex form, ππ(π₯π₯) = ππ(π₯π₯ β β)2 + ππ, which is very useful for graphing as it provides information about the location, shape, and direction of the parabola.
In the second part of this chapter, we examine properties and graphs of quadratic functions, including basic transformations of these graphs. Finally, these properties are used in solving application problems, particularly problems involving optimization. In the last section of this chapter, we study how to solve polynomial and rational inequalities using sign analysis.
Q.1 Methods of Solving Quadratic Equations
As defined in section F4, a quadratic equation is a second-degree polynomial equation in one variable that can be written in standard form as
ππππππ + ππππ + ππ = ππ,
where ππ, ππ, and ππ are real numbers and ππ β ππ. Such equations can be solved in many different ways, as presented below.
Solving by Graphing
To solve a quadratic equation, for example π₯π₯2 + 2π₯π₯ β 3 = 0, we can consider its left side as a function ππ(π₯π₯) = π₯π₯2 + 2π₯π₯ β 3 and the right side as a function ππ(π₯π₯) = 0. To satisfy the original equation, both function values must be equal. After graphing both functions on the same grid, one can observe that this happens at points of intersection of the two graphs.
So the solutions to the original equation are the π₯π₯-coordinates of the intersection points of the two graphs. In our example, these are the ππ-intercepts or the roots of the function ππ(π₯π₯) = π₯π₯2 + 2π₯π₯ β 3, as indicated in Figure 1.1.
Thus, the solutions to π₯π₯2 + 2π₯π₯ β 3 = 0 are ππ = βππ and ππ = ππ.
Note: Notice that the graphing method, although visually appealing, is not always reliable. For example, the solutions to the equation 49π₯π₯2 β 4 = 0 are π₯π₯ = 2
7 and π₯π₯ = β2
7. Such numbers would be very hard to read from the graph.
Thus, the graphing method is advisable to use when searching for integral solutions or estimations of solutions.
To find exact solutions, we can use one of the algebraic methods presented below.
π₯π₯
ππ(ππ) = ππππ + ππππ β ππ
1
1
solutions = x-intercepts
β3
ππ(ππ) = ππ
Figure 1.1
351
Solving by Factoring Many quadratic equations can be solved by factoring and employing the zero-product property, as in section F4.
For example, the equation π₯π₯2 + 2π₯π₯ β 3 = 0 can be solved as follows:
(π₯π₯ + 3)(π₯π₯ β 1) = 0 so, by zero-product property,
π₯π₯ + 3 = 0 or π₯π₯ β 1 = 0, which gives us the solutions
ππ = βππ or ππ = ππ.
Solving by Using the Square Root Property
Quadratic equations of the form ππππππ + ππ = ππ can be solved by applying the square root property.
Square Root For any positive real number ππ, if ππππ = ππ, then ππ = Β±βππ. Property: This is because βπ₯π₯2 = |π₯π₯|. So, after applying the square root operator to both sides of the
equation π₯π₯2 = ππ, we have οΏ½π₯π₯2 = βππ |π₯π₯| = βππ π₯π₯ = Β±βππ
The Β±βππ is a shorter recording of two solutions: βππ and ββππ.
For example, the equation 49π₯π₯2 β 4 = 0 can be solved as follows:
49π₯π₯2 β 4 = 0
49π₯π₯2 = 4
π₯π₯2 = 449
οΏ½π₯π₯2 = οΏ½ 449
π₯π₯ = Β±οΏ½ 449
ππ = Β±ππππ
Note: Using the square root property is a common solving strategy for quadratic equations where one side is a perfect square of an unknown quantity and the other side is a constant number.
apply square root to both sides of
the equation
/ +4
/ Γ· 49
Here we use the square root property. Remember the Β± sign!
352
Solve by the Square Root Property
Solve each equation using the square root property. a. (π₯π₯ β 3)2 = 49 b. 2(3π₯π₯ β 6)2 β 54 = 0 a. Applying the square root property, we have
οΏ½(π₯π₯ β 3)2 = β49
π₯π₯ β 3 = Β±7
π₯π₯ = 3 Β± 7 so
ππ = ππππ or ππ = βππ
b. To solve 2(3π₯π₯ β 6)2 β 54 = 0, we isolate the perfect square first and then apply the square root property. So,
2(3π₯π₯ β 6)2 β 54 = 0
(3π₯π₯ β 6)2 =542
οΏ½(3π₯π₯ β 6)2 = β27
3π₯π₯ β 6 = Β±3β3
3π₯π₯ = 6 Β± 3β3
π₯π₯ =6 Β± 3β3
3
π₯π₯ =3οΏ½2 Β± β3οΏ½
3
π₯π₯ = 2 Β± β3
Thus, the solution set is οΏ½ππ β βππ, ππ + βπποΏ½.
Caution: To simplify expressions such as 6+3β33
, we factor the numerator first. The common errors to avoid are
6+3β33
= 9β33
= 3β3 or
6+3β33
= 6 + β3 or
6+3β33
= 2 + 3β3
Solution
incorrect order of operations
incorrect canceling
incorrect canceling
/ +54,Γ· 2
/ +6
/ Γ· 3
/ +3
353 Solving by Completing the Square
So far, we have seen how to solve quadratic equations, πππ₯π₯2 + πππ₯π₯ + ππ = 0, if the expression πππ₯π₯2 + πππ₯π₯ + ππ is factorable or if the coefficient ππ is equal to zero. To solve other quadratic equations, we may try to rewrite the variable terms in the form of a perfect square, so that the resulting equation can already be solved by the square root property.
For example, to solve π₯π₯2 + 6π₯π₯ β 3 = 0, we observe that the variable terms π₯π₯2 + 6π₯π₯ could be written in perfect square form if we add 9, as illustrated in Figure 1.2. This is because
π₯π₯2 + 6π₯π₯ + 9 = (π₯π₯ + 3)2
Since the original equation can only be changed to an equivalent form, if we add 9, we must subtract 9 as well. (Alternatively, we could add 9 to both sides of the equation.) So, the equation can be transformed as follows:
π₯π₯2 + 6π₯π₯ β 3 = 0
π₯π₯2 + 6π₯π₯ + 9οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππππππππππππππ π π π π π π π π ππππ
β 9 β 3 = 0
(π₯π₯ + 3)2 = 12
οΏ½(π₯π₯ + 3)2 = β12
π₯π₯ + 3 = Β±2β3
π₯π₯ = βππΒ± ππβππ
Generally, to complete the square for the first two terms of the equation
ππππ + ππππ + ππ = ππ,
we take half of the π₯π₯-coefficient, which is ππππ, and square it. Then, we add and subtract that
number, οΏ½ππ2οΏ½2. (Alternatively, we could add οΏ½ππ
2οΏ½2 to both sides of the equation.)
This way, we produce an equivalent equation
ππππ + ππππ + οΏ½πππποΏ½ππβ οΏ½ππ
πποΏ½ππ
+ ππ = ππ, and consequently,
οΏ½ππ + πππποΏ½ππβ ππππ
ππ+ ππ = ππ.
observe that 3 comes from taking half of 6
/ +12
square root property
Completing the Square Procedure
Figure 1.2
π₯π₯2 π₯π₯ π₯π₯ π₯π₯ π₯π₯ π₯π₯ π₯π₯
π₯π₯2 π₯π₯ π₯π₯ π₯π₯
π₯π₯ π₯π₯ 1 1
1 1 1 1
π₯π₯ + 3 οΏ½β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―οΏ½
οΏ½β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―οΏ½
π₯π₯ +
3 π₯π₯ 1 1 1
take away 9
We can write this equation directly, by following the rule:
Write the sum of ππ and half of the middle coefficient, square the binomial, and subtract the perfect square of
the constant appearing in the bracket.
354
To complete the square for the first two terms of a quadratic equation with a leading coefficient of ππ β 1,
ππππππ + ππππ + ππ = ππ, we
divide the equation by ππ (alternatively, we could factor a out of the first two terms) so that the leading coefficient is 1, and then
complete the square as in the previous case, where ππ = ππ.
So, after division by ππ, we obtain
ππππ + ππππππ + ππ
ππ= ππ.
Since half of ππππ is ππ
ππππ, then we complete the square as follows:
οΏ½ππ + πππππποΏ½ππβ ππππ
ππππππ+ ππ
ππ= ππ.
Solve by Completing the Square
Solve each equation using the completing the square method.
a. π₯π₯2 + 5π₯π₯ β 1 = 0 b. 3π₯π₯2 β 12π₯π₯ β 5 = 0 a. First, we complete the square for π₯π₯2 + 5π₯π₯ by adding and subtracting οΏ½52οΏ½
2and then we apply the square root property. So, we have
π₯π₯2 + 5π₯π₯ + οΏ½52οΏ½2
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππππππππππππππ π π π π π π π π ππππ
β οΏ½52οΏ½2β 1 = 0
οΏ½π₯π₯ + 52οΏ½2β 25
4β 1 β 4
4= 0
οΏ½π₯π₯ + 52οΏ½2β 29
4= 0
οΏ½π₯π₯ + 52οΏ½2
= 294
π₯π₯ + 52
= Β±οΏ½294
Solution
Remember to subtract the perfect square of the constant appearing in
the bracket!
/ +294
apply square root to both sides of
the equation
remember to use
the Β± sign!
355
π₯π₯ + 52
= Β±β292
π₯π₯ =β5 Β± β29
2
Thus, the solution set is οΏ½βππββππππππ
, βππ+βππππππ
οΏ½.
Note: Unless specified otherwise, we are expected to state the exact solutions rather than their calculator approximations. Sometimes, however, especially when solving application problems, we may need to use a calculator to approximate the solutions. The reader is encouraged to check that the two decimal approximations of the above solutions are
βππββππππππ
β βππ.ππππ and βππ+βππππππ
β ππ.ππππ
b. In order to apply the strategy as in the previous example, we divide the equation by the leading coefficient, 3. So, we obtain
3π₯π₯2 β 12π₯π₯ β 5 = 0
π₯π₯2 β 4π₯π₯ β 53
= 0
Then, to complete the square for π₯π₯2 β 4π₯π₯, we may add and subtract 4. This allows us to rewrite the equation equivalently, with the variable part in perfect square form.
(π₯π₯ β 2)2 β 4 β 53 = 0
(π₯π₯ β 2)2 = 4 β 33
+ 53
(π₯π₯ β 2)2 = 173
π₯π₯ β 2 = Β±οΏ½173
π₯π₯ = ππ Β± βππππβππ
Note: The final answer could be written as a single fraction as shown below:
π₯π₯ = 2β3Β±β17β3
β β3β3
= ππΒ±βππππππ
Solving with Quadratic Formula
Applying the completing the square procedure to the quadratic equation
πππ₯π₯2 + πππ₯π₯ + ππ = 0,
with real coefficients ππ β 0, ππ, and ππ, allows us to develop a general formula for finding the solution(s) to any such equation.
/ β52
/ Γ· 3
356
Quadratic The solution(s) to the equation πππ₯π₯2 + πππ₯π₯ + ππ = 0, where ππ β 0, ππ, ππ are real coefficients, Formula are given by the formula
ππππ,ππ =βππ Β± βππππ β ππππππ
ππππ
Here ππππ,ππ denotes the two solutions, ππππ = βππ + οΏ½ππππβππππππππππ
, and ππππ = βππ β οΏ½ππππβππππππππππ
.
First, since ππ β 0, we can divide the equation πππ₯π₯2 + πππ₯π₯ + ππ = 0 by ππ. So, the equation to solve is
π₯π₯2 +πππππ₯π₯ +
ππππ
= 0
Then, we complete the square for π₯π₯2 + πππ π π₯π₯ by adding and subtracting the perfect square of
half of the middle coefficient, οΏ½ ππ2π π οΏ½2. So, we obtain
π₯π₯2 +πππππ₯π₯ + οΏ½
ππ2πποΏ½2
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππππππππππππππ π π π π π π π π ππππ
β οΏ½ππ
2πποΏ½2
+ππππ
= 0
οΏ½π₯π₯ +ππ
2πποΏ½2
β οΏ½ππ
2πποΏ½2
+ππππ
= 0
οΏ½π₯π₯ +ππ
2πποΏ½2
=ππ2
4ππ2βππππβ
4ππ4ππ
οΏ½π₯π₯ +ππ
2πποΏ½2
=ππ2 β 4ππππ
4ππ2
π₯π₯ +ππ
2ππ= Β±οΏ½
ππ2 β 4ππππ4ππ2
π₯π₯ +ππ
2ππ= Β±
βππ2 β 4ππππ2ππ
and finaly,
ππ =βππΒ± βππππ β ππππππ
ππππ,
which concludes the proof.
Solving Quadratic Equations with the Use of the Quadratic Formula
Using the Quadratic Formula, solve each equation, if possible. Then visualize the solutions graphically.
a. 2π₯π₯2 + 3π₯π₯ β 20 = 0 b. 3π₯π₯2 β 4 = 2π₯π₯ c. π₯π₯2 β β2π₯π₯ + 3 = 0
Proof:
QUADRATIC FORMULA
ππππ,ππ =βππΒ± βππππ β ππππππ
ππππ
/ βππ
2ππ
/ + οΏ½ππ
2πποΏ½2
,βππππ
357
a. To apply the quadratic formula, first, we identify the values of ππ, ππ, and ππ. Since the equation is in standard form, ππ = 2, ππ = 3, and ππ = β20. The solutions are equal to
π₯π₯1,2 =βππ Β± βππ2 β 4ππππ
2ππ=β3 Β± οΏ½32 β 4 β 2(β20)
2 β 2=β3 Β± β9 + 160
4
=β3 Β± 13
4=
β©βͺβ¨
βͺβ§β3 + 13
4=
104
=ππππ
β3 β 134
=β16
4= βππ
Thus, the solution set is οΏ½βππ, πππποΏ½.
These solutions can be seen as π₯π₯-intercepts of the function ππ(π₯π₯) = 2π₯π₯2 + 3π₯π₯ β 20, as
shown in Figure 1.3.
b. Before we identify the values of ππ, ππ, and ππ, we need to write the given equation 3π₯π₯2 β4 = 2π₯π₯ in standard form. After subtracting 4π₯π₯ from both sides of the given equation, we obtain
3π₯π₯2 β 2π₯π₯ β 4 = 0
Since ππ = 3, ππ = β2, and ππ = β5, we evaluate the quadratic formula,
π₯π₯1,2 =βππ Β± βππ2 β 4ππππ
2ππ=
2 Β± οΏ½(β2)2 β 4 β 3(β4)2 β 3
=2 Β± β4 + 48
6=
2 Β± β526
=2 Β± β4 β 13
6=
2 Β± 2β136
=2οΏ½1 Β± β13οΏ½
6=ππ Β± βππππ
ππ
So, the solution set is οΏ½ππββππππππ
, ππ+βππππππ
οΏ½. We may visualize solutions to the original equation, 3π₯π₯2 β 4 = 2π₯π₯, by graphing
functions ππ(π₯π₯) = 3π₯π₯2 β 4 and ππ(π₯π₯) = 2π₯π₯. The π₯π₯-coordinates of the intersection points are the solutions to the equation ππ(π₯π₯) = ππ(π₯π₯), and consequently to the original
equation. As indicated in Figure 1.4, the approximations of these solutions are 1ββ133
β
β0.87 and 1+β133
β 1.54. c. Substituting ππ = 1, ππ = ββ2, and ππ = 3 into the Quadratic Formula, we obtain
π₯π₯1,2 =β2 Β± οΏ½οΏ½ββ2οΏ½
2β 4 β 1 β 3
2 β 1=β2 Β± β4 β 12
2=β2 Β± ββ8
2
Since a square root of a negative number is not a real value, we have no real solutions.
Thus the solution set to this equation is β . In a graphical representation, this means that the graph of the function ππ(π₯π₯) = π₯π₯2 β β2π₯π₯ + 3 does not cross the x-axis. See Figure 1.5.
Solution
simplify by factoring
π₯π₯
ππ(ππ) = ππππππ + ππππ β ππππ
4
β4 2
solutions = x-intercepts
Figure 1.3
Figure 1.4
π₯π₯
ππ(ππ) = ππππππ β ππ
1
1
solutions = x-coordinates
of the intercepts
ππ(ππ) = ππππ
~1.54 ~ β 0.87
not a real number!
π₯π₯
ππ(ππ) = ππππ β βππππ + ππ
1
1
no solutions
Figure 1.5
358
Observation: Notice that we could find the solution set in Example 3c just by evaluating the radicand ππ2 β 4ππππ. Since this radicand was negative, we concluded that there is no solution to the given equation as a root of a negative number is not a real number. There was no need to evaluate the whole expression of Quadratic Formula.
So, the radicand in the Quadratic Formula carries important information about the number and nature of roots. Because of it, this radicand earned a special name, the discriminant.
Definition 1.1 The radicand ππππ β ππππππ in the Quadratic Formula is called the discriminant and it is denoted by β.
Notice that in terms of β, the Quadratic Formula takes the form
Observing the behaviour of the expression ββ allows us to classify the number and type of solutions (roots) of a quadratic equation with rational coefficients.
Characteristics of Roots (Solutions) Depending on the Discriminant
Suppose πππ₯π₯2 + πππ₯π₯ + ππ = 0 has rational coefficients ππ β 0, ππ, ππ, and β = ππππ β ππππππ. If β < ππ, then the equation has no real solutions, as οΏ½ππππππππππππππππ is not a real number.
If β = ππ, then the equation has one rational solution, βππππππ
.
If β > ππ, then the equation has two solutions, βππβββππππ
and βππ+ββππππ
. These solutions are - irrational, if β is not a perfect square number - rational, if β is a perfect square number (as οΏ½ππππππππππππππ π π π π π π ππππππ = ππππππππππππππ) In addition, if β β₯ ππ is a perfect square number, then the equation could be solved
by factoring.
Determining the Number and Type of Solutions of a Quadratic Equation
Using the discriminant, determine the number and type of solutions of each equation without solving the equation. If the equation can be solved by factoring, show the factored form of the trinomial.
a. 2π₯π₯2 + 7π₯π₯ β 15 = 0 b. 4π₯π₯2 β 12π₯π₯ + 9 = 0
c. 3π₯π₯2 β π₯π₯ + 1 = 0 d. 2π₯π₯2 β 7π₯π₯ + 2 = 0
ππππ,ππ =βππ Β± ββ
ππππ
359
a. β = 72 β 4 β 2 β (β15) = 49 + 120 = 169
Since 169 is a perfect square number, the equation has two rational solutions and it can be solved by factoring. Indeed, 2π₯π₯2 + 7π₯π₯ β 15 = (2π₯π₯ β 3)(π₯π₯ + 5).
b. β = (β12)2 β 4 β 4 β 9 = 144 β 144 = 0
β = 0 indicates that the equation has one rational solution and it can be solved by factoring. Indeed, the expression 4π₯π₯2 β 12π₯π₯ + 9 is a perfect square, (2π₯π₯ β 3)2.
c. β = (β1)2 β 4 β 3 β 1 = 1 β 12 = β11
Since β < 0, the equation has no real solutions and therefore it can not be solved by factoring.
d. β = (β7)2 β 4 β 2 β 2 = 49 β 16 = 33
Since β > 0 but it is not a perfect square number, the equation has two real solutions but it cannot be solved by factoring.
Solving Equations Equivalent to Quadratic
Solve each equation.
a. 2 + 7π₯π₯
= 5π₯π₯2
b. 2π₯π₯2 = (π₯π₯ + 2)(π₯π₯ β 1) + 1
a. This is a rational equation, with the set of β β {0} as its domain. To solve it, we
multiply the equation by the πΏπΏπΏπΏπΏπΏ = π₯π₯2. This brings us to a quadratic equation
2π₯π₯2 + 7π₯π₯ = 5 or equivalently
2π₯π₯2 + 7π₯π₯ β 5 = 0,
which can be solved by following the Quadratic Formula for ππ = 2, ππ = 7, and ππ =β5. So, we have
π₯π₯1,2 =β7 Β± οΏ½(β7)2 β 4 β 2(β5)
2 β 2=β7 Β± β49 + 40
4=βππΒ± βππππ
ππ
Since both solutions are in the domain, the solution set is οΏ½βππββππππππ
, βππ+βππππππ
οΏ½
b. To solve 1 β 2π₯π₯2 = (π₯π₯ + 2)(π₯π₯ β 1), we simplify the equation first and rewrite it in standard form. So, we have
1 β 2π₯π₯2 = π₯π₯2 + π₯π₯ β 2
β3π₯π₯2 β π₯π₯ + 3 = 0
Solution
/ βπ₯π₯, +2
/ β (β1)
Solution
360
3π₯π₯2 + π₯π₯ β 3 = 0
Since the left side of this equation is not factorable, we may use the Quadratic Formula. So, the solutions are
π₯π₯1,2 =β1 Β± οΏ½12 β 4 β 3(β3)
2 β 3=
1 Β± β1 + 366
=ππ Β± βππππ
ππ.
Q.1 Exercises
Concept Check True or False.
1. A quadratic equation is an equation that can be written in the form πππ₯π₯2 + πππ₯π₯ + ππ = 0, where ππ, ππ, and ππ are any real numbers.
2. If the graph of ππ(π₯π₯) = πππ₯π₯2 + πππ₯π₯ + ππ intersects the π₯π₯-axis twice, the equation πππ₯π₯2 + πππ₯π₯ + ππ = 0 has two solutions.
3. If the equation πππ₯π₯2 + πππ₯π₯ + ππ = 0 has no solution, the graph of ππ(π₯π₯) = πππ₯π₯2 + πππ₯π₯ + ππ does not intersect the π₯π₯-axis.
4. The Quadratic Formula cannot be used to solve the equation π₯π₯2 β 5 = 0 because the equation does not contain a linear term.
5. The solution set for the equation π₯π₯2 = 16 is {4}.
6. To complete the square for π₯π₯2 + πππ₯π₯, we add οΏ½ππ2οΏ½2.
7. If the discriminant is positive, the equation can be solved by factoring. Concept Check
For each function ππ, a) graph ππ(π₯π₯) using a table of values; b) find the x-intercepts of the graph; c) solve the equation ππ(π₯π₯) = 0 by factoring and compare these solutions to the x-intercepts of the graph.
8. ππ(π₯π₯) = βπ₯π₯2 β 3π₯π₯ + 2 9. ππ(π₯π₯) = π₯π₯2 + 2π₯π₯ β 3 10. ππ(π₯π₯) = 3π₯π₯ + π₯π₯(π₯π₯ β 2)
11. ππ(π₯π₯) = 2π₯π₯ β π₯π₯(π₯π₯ β 3) 12. ππ(π₯π₯) = 4π₯π₯2 β 4π₯π₯ β 3 13. ππ(π₯π₯) = β12
(2π₯π₯2 + 5π₯π₯ β 12) Solve each equation using the square root property.
14. π₯π₯2 = 49 15. π₯π₯2 = 32 16. ππ2 β 50 = 0
17. ππ2 β 24 = 0 18. 3π₯π₯2 β 72 = 0 19. 5π¦π¦2 β 200 = 0
361
20. (π₯π₯ β 4)2 = 64 21. (π₯π₯ + 3)2 = 16 22. (3ππ β 1)2 = 7
23. (5ππ + 2)2 = 12 24. π₯π₯2 β 10π₯π₯ + 25 = 45 25. π¦π¦2 + 8π¦π¦ + 16 = 44
26. 4ππ2 + 12ππ + 9 = 32 27. 25(π¦π¦ β 10)2 = 36 28. 16(π₯π₯ + 4)2 = 81
29. (4π₯π₯ + 3)2 = β25 30. (3ππ β 2)(3ππ + 2) = β5 31. 2π₯π₯ β 1 = 182π₯π₯β1
Solve each equation using the completing the square procedure.
32. π₯π₯2 + 12π₯π₯ = 0 33. π¦π¦2 β 3π¦π¦ = 0 34. π₯π₯2 β 8π₯π₯ + 2 = 0
35. ππ2 + 7ππ = 3ππ β 4 36. ππ2 β 4ππ = 4ππ β 16 37. π¦π¦2 + 7π¦π¦ β 1 = 0
38. 2π₯π₯2 β 8π₯π₯ = β4 39. 3ππ2 + 6ππ = β9 40. 3π¦π¦2 β 9π¦π¦ + 15 = 0
41. 5π₯π₯2 β 60π₯π₯ + 80 = 0 42. 2ππ2 + 6ππ β 10 = 0 43. 3π₯π₯2 + 2π₯π₯ β 2 = 0
44. 2π₯π₯2 β 16π₯π₯ + 25 = 0 45. 9π₯π₯2 β 24π₯π₯ = β13 46. 25ππ2 β 20ππ = 1
47. π₯π₯2 β 43π₯π₯ = β1
9 48. π₯π₯2 + 5
2π₯π₯ = β1 49. π₯π₯2 β 2
5π₯π₯ β 3 = 0
Given ππ(π₯π₯) and ππ(π₯π₯), find all values of π₯π₯ for which ππ(π₯π₯) = ππ(π₯π₯).
50. ππ(π₯π₯) = π₯π₯2 β 9 and ππ(π₯π₯) = 4π₯π₯ β 6 51. ππ(π₯π₯) = 2π₯π₯2 β 5π₯π₯ and ππ(π₯π₯) = βπ₯π₯ + 14
Discussion Point
52. Explain the errors in the following solutions of the equation 5π₯π₯2 β 8π₯π₯ + 2 = 0:
a. π₯π₯ = 8 Β± ββ82β4β5β22β8
= 8 Β± β64β4016
= 8 Β± β2416
= 8 Β± 2β616
= ππππ
Β± ππβππ
b. π₯π₯ = 8 Β± οΏ½(β8)2β4β5β22β8
= 8 Β± β64β4016
= 8 Β± β2416
= 8 Β± 2β616
= οΏ½10β616
= ππβππππ
6β616
= ππβππππ
Solve each equation with the aid of the Quadratic Formula, if possible. Illustrate your solutions graphically, using a table of values or a graphing utility.
53. π₯π₯2 + 3π₯π₯ + 2 = 0 54. π¦π¦2 β 2 = π¦π¦ 55. π₯π₯2 + π₯π₯ = β3
56. 2π¦π¦2 + 3π¦π¦ = β2 57. π₯π₯2 β 8π₯π₯ + 16 = 0 58. 4ππ2 + 1 = 4ππ
Solve each equation with the aid of the Quadratic Formula. Give the exact and approximate solutions up to two decimal places.
59. ππ2 β 4 = 2ππ 60. 2 β 2π₯π₯ = 3π₯π₯2 61. 0.2π₯π₯2 + π₯π₯ + 0.7 = 0
62. 2ππ2 β 4ππ + 2 = β3 63. π¦π¦2 + π¦π¦3
= 16 64. π₯π₯2
4β π₯π₯
2= 1
65. 5π₯π₯2 = 17π₯π₯ β 2 66. 15π¦π¦ = 2π¦π¦2 + 16 67. 6π₯π₯2 β 8π₯π₯ = 2π₯π₯ β3
ππ =βππ Β± βππππ β ππππππ
ππππ
362
Concept Check Use the discriminant to determine the number and type of solutions for each equation. Also, decide whether the equation can be solved by factoring or whether the quadratic formula should be used. Do not actually solve.
68. 3π₯π₯2 β 5π₯π₯ β 2 = 0 69. 4π₯π₯2 = 4π₯π₯ + 3 70. π₯π₯2 + 3 = β2β3π₯π₯
71. 4π¦π¦2 β 28π¦π¦ + 49 = 0 72. 3π¦π¦2 β 10π¦π¦ + 15 = 0 73. 9π₯π₯2 + 6π₯π₯ = β1
Find the value(s) of the constant k, so that each equation will have exactly one rational solution.
74. π₯π₯2 + πππ¦π¦ + 49 = 0 75. 9π¦π¦2 β 30π¦π¦ + ππ = 0 76. πππ₯π₯2 + 8π₯π₯ + 1 = 0
Discussion Point
77. Is it possible for the solution of a quadratic equation with integral coefficients to include only one irrational number? Why or why not?
Solve each equation using any algebraic method. State the solutions in their exact form .
78. β2π₯π₯(π₯π₯ + 2) = β3 79. (π₯π₯ + 2)(π₯π₯ β 4) = 1 80. (π₯π₯ + 2)(π₯π₯ + 6) = 8
81. (2π₯π₯ β 3)2 = 8(π₯π₯ + 1) 82. (3π₯π₯ + 1)2 = 2(1 β 3π₯π₯) 83. 2π₯π₯2 β (π₯π₯ + 2)(π₯π₯ β 3) = 12
84. (π₯π₯ β 2)2 + (π₯π₯ + 1)2 = 0 85. 1 + 2π₯π₯
+ 5π₯π₯2
= 0 86. π₯π₯ = 2(π₯π₯+3)π₯π₯+5
87. 2 + 1π₯π₯
= 3π₯π₯2
88. 3π₯π₯
+ π₯π₯3
= 52 89. 1
π₯π₯+ 1
π₯π₯+4= 1
7