1. § 5.7 Polynomial Equations and Their Applications

2. Blitzer, Intermediate Algebra, 4e – Slide #91 Solving Polynomial Equations Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the standard form where a, b, and c are real numbers, with. A quadratic equation in x is also called a second-degree polynomial equation in x. The Zero-Product Rule If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. If AB = 0, then A = 0 or B = 0.

3. Blitzer, Intermediate Algebra, 4e – Slide #92 Solving Polynomial Equations Solving a Quadratic Equation by Factoring 1)If necessary, rewrite the equation in the standard form, moving all terms to one side, thereby obtaining zero on the other side. 2) Factor completely. 3) Apply the zero-product principle, setting each factor containing a variable equal to zero. 4) Solve the equations in step 3. 5) Check the solutions in the original equation.

4. Blitzer, Intermediate Algebra, 4e – Slide #93 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 1) Move all terms to one side and obtain zero on the other side. Subtract 45 from both sides and write the equation in standard form. Subtract 45 from both sides Simplify 2) Factor. Factor

5. Blitzer, Intermediate Algebra, 4e – Slide #94 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. Check 9: Check -5: ? ? ? ?

6. Blitzer, Intermediate Algebra, 4e – Slide #95 Solving Polynomial EquationsCONTINUED Check 9: Check -5: The solutions are 9 and -5. The solution set is {9,-5}. ?? The graph of lies to the right. true

7. Blitzer, Intermediate Algebra, 4e – Slide #96 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 1) Move all terms to one side and obtain zero on the other side. Subtract 4x from both sides and write the equation in standard form. NOTE: DO NOT DIVIDE BOTH SIDES BY x. WE WOULD LOSE A POTENTIAL SOLUTION!!! Subtract 4x from both sides Simplify 2) Factor. Factor

8. Blitzer, Intermediate Algebra, 4e – Slide #97 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. Check 0: Check 4: ?? true

9. Blitzer, Intermediate Algebra, 4e – Slide #98 Solving Polynomial EquationsCONTINUED The solutions are 0 and 4. The solution set is {0,4}. The graph of lies to the right.

10. Blitzer, Intermediate Algebra, 4e – Slide #99 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: Be careful! Although the left side of the original equation is factored, we cannot use the zero-product principle since the right side of the equation is NOT ZERO!! 1) Move all terms to one side and obtain zero on the other side. Subtract 14 from both sides and write the equation in standard form. Simplify Subtract 14 from both sides

11. Blitzer, Intermediate Algebra, 4e – Slide #100 Solving Polynomial EquationsCONTINUED 2) Factor. Before we can factor the equation, we must simplify it first. FOIL Simplify Now we can factor the polynomial equation.

12. Blitzer, Intermediate Algebra, 4e – Slide #101 Solving Polynomial EquationsCONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. or 5) Check the solutions in the original equation. Check 3: Check -6: ?? ??

13. Blitzer, Intermediate Algebra, 4e – Slide #102 Solving Polynomial EquationsCONTINUED Check 3: Check -6: true The solutions are 3 and -6. The solution set is {3,-6}. The graph of lies to the right.

14. Blitzer, Intermediate Algebra, 4e – Slide #103 Solving Polynomial EquationsEXAMPLE SOLUTION Solve by factoring: 1) Move all terms to one side and obtain zero on the other side. This is already done. + 2) Factor. Use factoring by grouping. Group terms that have a common factor. Common factor is Common factor is -1.

15. Blitzer, Intermediate Algebra, 4e – Slide #104 Solving Polynomial EquationsCONTINUED Factor from the first two terms and -1 from the last two terms Factor out the common binomial, x – 2, from each term Factor completely by factoring as the difference of two squares

16. Blitzer, Intermediate Algebra, 4e – Slide #105 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. CONTINUED or 5) Check the solutions in the original equation. Check the three solutions 2, -1, and 1, by substituting them into the original equation. Can you verify that the solutions are 2, -1, and 1? The graph of lies to the right.

17. Blitzer, Intermediate Algebra, 4e – Slide #106 Polynomial Equations in ApplicationEXAMPLE A gymnast dismounts the uneven parallel bars at a height of 8 feet with an initial upward velocity of 8 feet per second. The function describes the height of the gymnast’s feet above the ground, s (t), in feet, t seconds after dismounting. The graph of the function is shown below.

18. Blitzer, Intermediate Algebra, 4e – Slide #107 Polynomial Equations in ApplicationSOLUTION When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph. We note that the graph of the equation passes through the line y = 8 twice. Once when x = 0 and once when x = 0.5. This can be verified by determining when y = s (t) = 8. That is, CONTINUED Original equation Replace s (t) with 8 Subtract 8 from both sides Factor

19. Blitzer, Intermediate Algebra, 4e – Slide #108 Polynomial Equations in Application Now we set each factor equal to zero. CONTINUED We have just verified the information we deduced from the graph. That is, the gymnast will indeed be 8 feet off the ground at t = 0 seconds and at t = 0.5 seconds. These solutions are identified by the dots on the graph on the next page. or

20. Blitzer, Intermediate Algebra, 4e – Slide #109 Polynomial Equations in ApplicationCONTINUED

21. Blitzer, Intermediate Algebra, 4e – Slide #110 The Pythagorean Theorem The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse. If the legs have lengths a and b, and the hypotenuse has length c, then A C B c a b Hypotenuse Leg

22. Blitzer, Intermediate Algebra, 4e – Slide #111 The Pythagorean TheoremEXAMPLE SOLUTION A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire. Let’s begin with a picture. Tree 5 feet

23. Blitzer, Intermediate Algebra, 4e – Slide #112 The Pythagorean Theorem Since the wire is 1 foot longer than the height that it reaches on the tree, if we call the length of the wire (the quantity we wish to determine) x, then the height of the tree would be x -1. Tree 5 feet CONTINUED x-1 x

24. Blitzer, Intermediate Algebra, 4e – Slide #113 The Pythagorean Theorem We can now use the Pythagorean Theorem to solve for x, the length of the wire. CONTINUED This is the equation arising from the Pythagorean Theorem Square x – 1 and 5 Add 1 and 25 Subtract from both sides Add 2x to both sides Divide both sides by 2 Therefore, the solution is x = 13 feet.