1. Copyright © 2010 Pearson Education, Inc
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 1- 1

2. Quadratic Functions and Equations11
Quadratic Functions and Equations
11.1 Quadratic Equations
11.2 The Quadratic Formula
11.3 Studying Solutions of Quadratic Equations
11.4 Applications Involving Quadratic Equations
11.5 Equations Reducible to Quadratic
11.6 Quadratic Functions and Their Graphs
11.7 More About Graphing Quadratic Functions
11.8 Problem Solving and Quadratic Functions
11.9 Polynomial and Rational Inequalities
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3. The Principle of Square Roots Completing the Square Problem Solving
Quadratic Equations
11.1
The Principle of Square Roots
Completing the Square
Problem Solving
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4. The Principle of Square Roots
Let’s consider x2 = 25. We know that the number 25 has two real-number square roots, 5 and −5, the solutions of the equation. Thus we see that square roots can provide quick solutions for equations of the type x2 = k.
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5. The Principle of Square Roots
For any real number k, if x2 = k, then
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6. Example
Solve 5x 2 = 15. Give exact solutions and approximations to three decimal places.
Solution
Isolating x2
Using the principle of square roots
We often use the symbol to represent both solutions.
The solutions are which round to and – The check is left to the student.
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7. Example Solution Solve 16x2 + 9 = 0.
Recall that
The solutions are The check is left to the student.
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8. The Principle of Square Roots (Generalized Form)
For any real number k and any algebraic expression X:
If X 2 = k, then
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9. Example Solution Solve: (x + 3)2 = 7. The solutions are
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10. Completing the Square
Not all quadratic equations can be solved as in the previous examples. By using a method called completing the square, we can use the principle of square roots to solve any quadratic equation.
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11. Example Solution x2 + 10x + 25 = –4 + 25 Solve x2 + 10x + 4 = 0.
Adding 25 to both sides.
x2 + 10x + 25 = –4 + 25
Factoring
Using the principle of square roots
The solutions are
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12. To Solve a Quadratic Equation in x by Completing the Square
1. Isolate the terms with variables on one side of the equation, and arrange them in descending order.
2. Divide both sides by the coefficient of x2 if that coefficient is not 1.
3. Complete the square by taking half of the coefficient of x and adding its square to both sides.
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13. 6. Solve for x by adding or subtracting on both sides.
4. Express the trinomial as the square of a binomial (factor the trinomial) and simplify the other side.
5. Use the principle of square roots (find the square roots of both sides).
6. Solve for x by adding or subtracting on both sides.
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14. Problem Solving
After one year, an amount of money P, invested at 4% per year, is worth 104% of P, or P(1.04). If that amount continues to earn 4% interest per year, after the second year the investment will be worth 104% of P(1.04), or P(1.04)2. This is called compounding interest since after the first period, interest is earned on both the initial investment and the interest from the first period. Generalizing, we have the following.
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15. The Compound Interest Formula
If the amount of money P is invested at interest rate r, compounded annually, then in t years, it will grow to the amount A given by
(r is written in decimal notation.)
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16. Example Solution 6765 = 5800(1 + r)2
Jackson invested $5800 at an interest rate of r, compounded annually. In 2 yr, it grew to $ What was the interest rate?
Solution
Familiarize. We are already familiar with the compound-interest formula.
2. Translate. The translation consists of substituting into the formula:
6765 = 5800(1 + r)2
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17. 5. State. The interest rate was 8%.
3. Carry out. Solve for r:
6765/5800 = (1 + r)2
4. Check. Since the interest rate cannot negative, we need only to check .08 or 8%. If $5800 were invested at 8% compounded annually, then in 2 yr it would grow to 5800(1.08)2, or $ The number 8% checks.
5. State. The interest rate was 8%.
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18. Solving Using the Quadratic Formula Approximating Solutions
11.2
Solving Using the Quadratic Formula
Approximating Solutions
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19. Solving Using the Quadratic Formula
Each time we solve by completing the square, the procedure is the same. When a procedure is repeated many times, a formula can often be developed to speed up our work.
If we begin with a quadratic equation in standard form, ax2 + bx + c = 0, and solve by completing the square we arrive at the quadratic formula.
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20. The Quadratic Formula The solutions of ax2 + bx + c = 0, are given by
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21. Example Solution Solve 3x2 + 5x = 2 using the quadratic formula.
First determine a, b, and c:
3x2 + 5x – 2 = 0;
a = 3, b = 5, and c = –2.
Substituting
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22. The solutions are 1/3 and –2. The check is left to the student.
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23. To Solve a Quadratic Equation
1. If the equation can easily be written in the form ax2 = p or (x + k)2 = d, use the principle of square roots.
2. If step (1) does not apply, write the equation in the form ax2 + bx + c = 0.
3. Try factoring using the principle of zero products.
4. If factoring seems difficult or impossible, use the quadratic formula. Completing the square can also be used.
The solutions of a quadratic equation can always be found using the quadratic formula. They cannot always be found by factoring.
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24. Recall that a second-degree polynomial in one variable is said to be quadratic. Similarly, a second-degree polynomial function in one variable is said to be a quadratic function.
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25. Example Solution Solve x2 + 7 = 2x using the quadratic formula.
First determine a, b, and c:
x2 – 2x + 7 = 0;
a = 1, b = –2, and c = 7.
Substituting
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26. The check is left to the student.
The solutions are
The check is left to the student.
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27. Approximating Solutions
When the solution of an equation is irrational, a rational-number approximation is often useful. This is often the case in real-world applications similar to those found in section 8.4.
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28. Example
Use a calculator to approximate
Solution
Take the time to familiarize yourself with your calculator:
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29. Studying Solutions of Quadratic Equations
11.3
The Discriminant
Writing Equations from Solutions
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30. The Discriminant
It is sometimes enough to know what type of number a solution will be, without actually solving the equation. From the quadratic formula, b2 – 4ac, is known as the discriminant. The discriminant determines what type of number the solutions of a quadratic equation are. The cases are summarized on the next slide.
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31. Discriminant b2 – 4ac Nature of Solutions
One solution; a rational number
Positive
Perfect square
Not a perfect square
Two different real-number solutions
Solutions are rational
Solutions are irrational conjugates
Negative
Two different imaginary-number solutions (complex conjugates)
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32. Example Solution b2 – 4ac = (–1)2 – 4(4)(1) = –15.
For the equation 4x2 – x + 1 = 0, determine what type of number the solutions are and how many solutions exist.
Solution
First determine a, b, and c: a = 4, b = –1, and c = 1.
Compute the discriminant:
b2 – 4ac = (–1)2 – 4(4)(1) = –15.
Since the discriminant is negative, there are two imaginary-number solutions that are complex conjugates of each other.
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33. Example Solution b2 – 4ac = (–10)2 – 4(5)(5) = 0.
For the equation 5x2 – 10x + 5 = 0, determine what type of number the solutions are and how many solutions exist.
Solution
First determine a, b, and c: a = 5, b = –10, and c = 5.
Compute the discriminant:
b2 – 4ac = (–10)2 – 4(5)(5) = 0.
There is exactly one solution, and it is rational. This indicates that 5x2 – 10x + 5 = 0 can be solved by factoring.
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34. Example Solution b2 – 4ac = (7)2 – 4(2)(–3) = 73.
For the equation 2x2 + 7x – 3 = 0, determine what type of number the solutions are and how many solutions exist.
Solution
First determine a, b, and c: a = 2, b = 7, and c = –3.
Compute the discriminant:
b2 – 4ac = (7)2 – 4(2)(–3) = 73.
The discriminant is a positive number that is not a perfect square. Thus there are two irrational solutions that are conjugates of each other.
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35. Writing Equations from Solutions
We know by the principle of zero products that (x – 1)(x + 4) = 0 has solutions 1 and -4. If we know the solutions of an equation, we can write an equation, using the principle in reverse.
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36. Example Solution Find an equation for which 5 and –4/3 are solutions.
x = 5 or x = –4/3
x – 5 = 0 or x + 4/3 = 0
Get 0’s on one side
Using the principle of zero products
(x – 5)(x + 4/3) = 0
x2 – 5x + 4/3x – 20/3 = 0
Multiplying
3x2 – 11x – 20 = 0
Combining like terms and clearing fractions
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37. Example Solution Find an equation for which 3i and –3i are solutions.
x = 3i or x = –3i
x – 3i = 0 or x + 3i = 0
Get 0’s on one side
Using the principle of zero products
(x – 3i)(x + 3i) = 0
x2 – 3ix + 3ix – 9i2 = 0
Multiplying
x = 0
Combining like terms
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38. Applications Involving Quadratic Equations
11.4
Solving Problems
Solving Formulas
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39. Solving Problems
As we found in Section 6.7, some problems translate to rational equations. The solution of such rational equations can involve quadratic equations.
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40. Example
Cade traveled 48 miles at a certain speed. If he had gone 4 mph faster, the trip would have taken 1 hr less. Find Cade’s average speed.
Solution
Familiarize. As in Section 6.5, we can create a table. Let r represent the rate, in miles per hour, and t the time, in hours for Cade’s trip.
Distance
Speed
Time 48 r t
r + 4
t – 1
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41. and 2. Translate. From the table we obtain
3. Carry out. A system of equations has been formed. We substitute for r from the first equation into the second and solve the resulting equation:
Multiplying by the LCD
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43. This is 1 hr less than the trip actually took, so the answer checks.
4. Check. Note that we solved for t, not r as required. Since negative time has no meaning here, we disregard the –3 and use 4 to find r:
12 mph.
To see if 12 mph checks, we increase the speed 4 mph to 16 and see how long the trip would have taken at that speed:
This is 1 hr less than the trip actually took, so the answer checks.
5. State. Cade traveled at 12 mph.
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44. Solving Formulas
Recall that to solve a formula for a certain letter, we use the principles for solving equations to get that letter alone on one side.
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45. Example Solve Solution Multiplying both sides by 2
Writing standard form
Using the quadratic formula
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46. To Solve a Formula for a Letter – Say, h
1. Clear fractions and use the principle of powers, as needed. Perform these steps until radicals containing h are gone and h is not in any denominator.
2. Combine all like terms.
3. If the only power of h is h1, the equation can be solved as in Sections 2.3 and 7.5.
4. If h2 appears but h does not, solve for h2 and use the principle of square roots to solve for h.
5. If there are terms containing both h and h2, put the equation in standard form and use the quadratic formula.
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47. Equations Reducible to Quadratic
11.5
Recognizing Equations in Quadratic Form
Radical Equations and Rational Equations
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48. Recognizing Equations in Quadratic Form
Certain equations that are not really quadratic can be thought of in such a way that they can be solved as quadratic. For example, because the square of x2 is x4, the equation x4 – 5x2 + 4 = 0 is said to be “quadratic in x2”:
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49. x4 – 5x2 + 4 = 0
(x2)2 – 5(x2) + 4 = 0
u2 – 5u = 0.
The last equation can be solved by factoring or by the quadratic formula. Then, remembering that u = x2, we can solve for x. Equations that can be solved like this are reducible to quadratic, or in quadratic form.
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50. Example Solve x4 – 5x2 + 4 = 0. Solution u2 – 5u + 4 = 0
Let u = x2. Then we solve by substituting u for x2 and u2 for x4:
u2 – 5u + 4 = 0
(u – 1)(u – 4) = 0
Factoring
Principle of zero products
u – 1 = 0 or u – 4 = 0
u = 1 or u = 4
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51. x2 = 1 or x2 = 4
Replace u with x2
To check, note that for both x = 1 and x = –1, we have x2 = 1 and x4 = 1. Similarly, for both x = 2 and x = –2, we have x2 = 4 and x4 = 16. Thus instead of making four checks, we need make only two.
Check:
x = 1:
x = 2:
x4 – 5x2 + 4 = 0
x4 – 5x2 + 4 = 0
(16) – 5(4) + 4 = 0
(1) – 5(1) + 4 = 0
TRUE
TRUE
The solutions are 1, –1, 2, and –2.
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52. Caution! A common error on problems like the previous example is to solve for u but forget to solve for x. Remember to solve for the original variable!
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53. Radical and Rational Equations
Sometimes rational equations, radical equations, or equations containing exponents that are fractions are reducible to quadratic. It is especially important that answers to these equations be checked in the original equation.
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54. Example Solve Solution u2 – 8u – 9 = 0 (u – 9)(u +1) = 0
Let u = Then we solve by substituting u for and u2 for x:
u2 – 8u – 9 = 0
(u – 9)(u +1) = 0
u – 9 = 0 or u + 1 = 0
u = 9 or u = –1
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55. Check: x = 81: x = 1: The solution is 81. FALSE TRUE
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56. Example Solve Solution u2 + 4u – 2 = 0
Let u = t −1. Then we solve by substituting u for t −1 and u2 for t −2:
u2 + 4u – 2 = 0
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57. The check is left to the student.
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58. To Solve an Equation That is Reducible to Quadratic
1. Look for two variable expressions in the equation. One expressions should be the square of the other.
2. Write down any substitutions that you are making.
3. Remember to solve for the variable that is used in the original equation.
4. Check possible answers in the original equation.
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59. Quadratic Functions and Their Graphs
11.6
The Graph of f (x) = ax2
The Graph of f (x) = a(x – h)2
The Graph of f (x) = a(x – h)2 + k
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60. The Graph of f (x) = ax2
All quadratic functions have graphs similar to y = x2. Such curves are called parabolas. They are U-shaped and symmetric with respect to a vertical line known as the parabola’s axis of symmetry. For the graph of f (x) = x2, the y-axis is the axis of symmetry. The point (0, 0) is known as the vertex of this parabola.
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61. Example Graph: Solution 1 –1 2 –2 8 (0, 0) (1, 2) (–1, 2) (2, 8)
(−2, 8)
(2, 8) x
f(x) = x2
(x, f(x)) 1 –1 2 –2 8
(0, 0)
(1, 2)
(–1, 2)
(2, 8)
(–2, 8)
(−1, 2)
(1, 2)
(0,0)
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62. Example Graph: Solution 1 –1 2 –2 –3 –12 (0, 0) (1, –3) (–1, –3)
f(x)
(x, f(x)) 1 –1 2 –2 –3
–12
(0, 0)
(1, –3)
(–1, –3)
(2, –12)
(–2, –12)
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63. 4 3 6 2 5 1
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64. Graphing f (x) = ax2
The graph of f (x) = ax2 is a parabola with x = 0 as
its axis of symmetry. Its vertex is the origin.
For a > 0, the parabola opens upward. For a < 0,
the parabola opens downward.
If |a| is greater than 1, the parabola is narrower than y = x2.
If |a| is between 0 and 1, the parabola is wider than y = x2.
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65. The Graph of f (x) = a(x – h)2
We could next consider graphs of f (x) = ax2 + bx + c, where b and c are not both 0. It turns out to be convenient to first graph f (x) = a(x – h)2, where h is some constant. This allows us to observe similarities to the graphs drawn in previous slides.
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66. Example Graph: Solution vertex 1 –1 2 3 4 9 x f(x) = (x – 2)21 –1 2 3 4 9
vertex
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67. Graphing f (x) = a(x – h)2
The graph of f (x) = a(x – h)2 has the same shape
as the graph of y = ax2.
If h is positive, the graph of y = ax2 is shifted h
units to the right.
If h is negative, the graph of y = ax2 is shifted |h| units to the left.
The vertex is (h, 0) and the axis of symmetry is
x = h.
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68. The Graph of f (x) = a(x – h)2 + k
Given a graph of f (x) = a(x – h)2, what happens if we add a constant k? Suppose that we add 2. This increases f (x) by 2, so the curve is moved up. If k is negative, the curve is moved down. The axis of symmetry for the parabola remains x = h, but the vertex will be at (h, k), or equivalently (h, f (h)).
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69. Graphing f (x) = a(x – h)2 + k
The graph of f (x) = a(x – h)2 + k has the same shape
as the graph of y = a(x – h)2.
If k is positive, the graph of y = a(x – h)2 is
shifted k units up.
If k is negative, the graph of y = a(x – h)2 is
shifted |k| units down.
The vertex is (h, k), and the axis of symmetry is x = h.
For a > 0, k is the minimum function value. For a < 0, the maximum function value is k.
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70. Example Graph: and find the maximum function value. Solution–1 –2 –3 –4 –5
-11/2
–3/2
Maximum = −1
vertex
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71. More About Graphing Quadratic Functions
11.7
Completing the Square
Finding Intercepts
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72. Completing the Square
By completing the square, we can rewrite any polynomial ax2 + bx + c in the form a(x – h)2 + k. Once that has been done, the procedures discussed in Section 11.6 will enable us to graph any quadratic function.
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73. Example Solution Graph: f (x) = x2 – 2x – 1 = (x2 – 2x) – 1y -3 2 -2 3 -1 1 6 5 4 -4 -5
= (x2 – 2x) – 1
= (x2 – 2x + 1 – 1) – 1
= (x2 – 2x + 1) – 1 – 1
= (x – 1)2 – 2
The vertex is at (1, –2).
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74. Example Solution Graph: f (x) = –2x2 + 6x – 3 = –2(x2 – 3x) – 3y -3 2 -2 3 -1 1 6 5 4 -4 -5
= –2(x2 – 3x) – 3
= –2(x2 – 3x + 9/4 – 9/4) – 3
= –2(x2 – 3x + 9/4) – /4
= –2(x – 3/2)2 + 3/2
The vertex is at (3/2, 3/2).
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75. The Vertex of a Parabola
The vertex of a parabola given by f (x) = ax2 + bx + c is
The x-coordinate of the vertex is –b/(2a).
The axis of symmetry is x = -b/(2a).
The second coordinate of the vertex is most commonly found by computing
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76. Finding Intercepts
For any function f, the y-intercept occurs at f (0). Thus for f (x) = ax2 + bx + c, the y-intercept is simply (0, c). To find x-intercepts, we look for points where y = 0 or f (x) = 0. Thus, for f (x) = ax2 + bx + c, the x-intercepts occur at those x-values for which
ax2 + bx + c = 0
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77. f (x) = ax2 + bx + c y - intercept x - intercepts y x 6 5 4 3 2 1 -1 -2 x -3 -4 -5
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78. Example
Find the x- and y-intercepts of the graph of f (x) = x2 + 3x – 1.
Solution
The y-intercept is simply (0, f (0)), or (0, –1). To find the x-intercepts, we solve the equation:
x2 + 3x – 1 = 0.
Since we are unable to solve by factoring, we use the quadratic formula to get
If graphing, we would approximate to get (–3.3, 0) and (0.3, 0).
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79. Problem Solving and Quadratic Functions
11.8
Maximum and Minimum Problems
Fitting Quadratic Functions to Data
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80. Maximum and Minimum Problems
We have seen that for any quadratic function f, the value of f (x) at the vertex is either a maximum or a minimum. Thus problems in which a quantity must be maximized or minimized can be solved by finding the coordinates of the vertex, assuming the problem can be modeled with a quadratic function.
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81. Example
A farmer has 200 ft of fence with which to form a rectangular pen on his farm. If an existing fence forms one side of the rectangle, what dimensions will maximize the size of the area?
Solution
Familiarize. We make a drawing and label it, letting w = the width of the rectangle, in feet and l = the length of the rectangle, in feet.
Existing fence w l
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82. Recall that Area = lw and Perimeter = 2w + 2l
Recall that Area = lw and Perimeter = 2w + 2l. Since the existing fence forms one length of the rectangle, the fence will comprise three sides. Thus
2w + l = 200.
2. Translate. We have two equations: One guarantees that all 200 ft of fence will be used; the other expresses area in terms of length and width.
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83. Factoring and completing the square, we get
3. Carryout. We need to express A as a function of l or w but not both. To do so, we solve for l in the first equation to obtain l = 200 – 2w. Substituting for l in the second equation, we get a quadratic function:
A = (200 – 2w)w
= 200w – 2w2
Factoring and completing the square, we get
A = –2(w – 50)
The maximum area, 5000 ft2, occurs when w = 50 ft and l = 200 – 2(50), or 100 ft.
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84. 4. Check. The check is left to the student.
5. State. The dimensions for the largest rectangular area for the pen that can be enclosed is 50 ft by 100 ft.
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85. Fitting Quadratic Functions to Data
Whenever a certain quadratic function fits a situation, that function can be determined if three inputs and their outputs are known. Each of the given ordered pairs is called a data point.
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86. Example
Use the data points (0, 10.4), (3, 16.8), and (6, 12.6) to find a quadratic function that fits the data.
Solution
We are looking for a function of the form f (x) = ax2 + bx + c, given that f (0) = 10.4, f (3) = 16.8, and f (6) = Thus
a(0)2 + b(0) + c = 10.4
a(3)2 + b(3) + c = 16.8
a(6)2 + b(6) + c = 12.6
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87. Simplifying we see that we need to solve the system
c = (1)
9a + 3b + c = (2)
36a + 6b + c = (3)
Substituting c = 10.4 into equations (2) and (3) and solving the resulting system we get
So we have
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88. Polynomial and Rational Inequalities
11.9
Quadratic and Other Polynomial Inequalities
Rational Inequalities
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89. Quadratic and Other Polynomial Inequalities
Inequalities like the following are called polynomial inequalities:
Second-degree polynomial inequalities in one variable are called quadratic inequalities. To solve polynomial inequalities, we often focus attention on where the outputs of a polynomial function are positive and where they are negative.
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90. Example Solve: Solution
Consider the related function f (x) = x2 + 2x – 8 and its graph. Since the graph opens upward, the y –values are positive outside the interval formed by the x-intercepts. x y -6 4 -4 6 -2 2 8 -8
-10
-14
-12
y = x2 + 2x – 8
positive y-values
Thus the solution set of the inequality is
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91. In the next example, we simplify the process by concentrating on the sign of a polynomial function over each interval formed by the x-intercepts. We will do this by tracking the sign of each factor. By looking at how many positive or negative factors are being multiplied, we will be able to determine the sign of the polynomial function.
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92. Example Solution We first solve the related equation: f (x) = 0
x3 – 9x = 0
x(x – 3)(x + 3) = 0
Factor
Using the principle of zero products
x = 0 or x – 3 = 0 or x + 3 = 0
x = 0 or x = 3 or x = –3.
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93. The function f has zeros at –3, 0 and 3
The function f has zeros at –3, 0 and 3. We will use the factorization f (x) = x(x – 3)(x + 3). The product x(x – 3)(x + 3) is positive or negative, depending on the signs of x, x – 3, and x This is easily determined using a chart.
Sign of x:
Sign of x – 3:
Sign of x + 3:
Sign of product
x(x – 3)(x + 3) –
Interval: + –
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94. A product is negative when it has an odd number of negative factors
A product is negative when it has an odd number of negative factors. Since the sign allows for equality, the endpoints –3, 0, and 3 are solutions. From the chart, we see that the solution set is
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95. To Solve a Polynomial Inequality Using Factors
1. Add or subtract to get 0 on one side and solve the related polynomial equation by factoring.
2. Use the numbers found in step (1) to divide the number line into intervals.
3. Using a test value from each interval, determine the sign of each factor over that interval. First find the sign of each factor, and then determine the sign of the product of the factors. Remember that the product of an odd number of negative numbers is negative.
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96. To Solve a Polynomial Inequality Using Factors
4. Select the interval(s) for which the inequality is satisfied and write set-builder notation or interval notation for the solution set. Include the endpoints of the intervals when is used.
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97. Rational Inequalities
Inequalities involving rational expressions are called rational inequalities. Like polynomial inequalities, rational inequalities can be solved using test values. Unlike polynomials, however, rational expressions often have values for which the expression is undefined.
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98. Example Solve: Solution
We write the related equation by changing the symbol to = :
Next we solve the related equation:
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99. In the case of rational inequalities, we must always find any values that make the denominator 0. As noted at the beginning that occurs when x = 3. Now we use 3 and 7 to divide the number line into intervals:
A B C
We test a number in each interval to see where the original inequality is satisfied:
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100. 0 is not a solution, so interval A is not part of the solution set.
A: Test 0,
4 is a solution, so interval B is part of the solution set.
B: Test 4,
10 is not a solution, so interval C is not part of the solution set.
C: Test 10,
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101. The solution set includes the interval B
The solution set includes the interval B. The endpoint 7 is included because the inequality symbol is and 7 is a solution of the related equation. The number 3 is not included because (x + 5)/(x – 3) is undefined for x = 3. This the solution set of the original inequality is
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102. To Solve a Rational Inequality
1.. Find any replacements for which the rational expression is undefined.
2. Change the inequality symbol to an equals sign and solve the related equation
3. Use the numbers found in step (1) and (2) to divide the number line into intervals.
4. Substitute a test value from each interval into the inequality. If the number is a solution, then the interval to which it belongs is part of the solution set.
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103. To Solve a Rational Inequality
5. Select the interval(s) and any endpoints for which the inequality is satisfied and write set-builder notation or interval notation for the solution set. If the inequality symbol is then the solutions from step (2) are also included in the solution set. Those numbers found in step (1) should be excluded from the solution set, even if they are solutions from step (2).
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