1.5 Applications of Quadratic Equations
- 1. 1.5 Applications of Quadratic Equations Chapter 1 Equations and Inequalities
- 2. Concepts & Objectives ⚫ Applications of Quadratic Equations ⚫ Geometry problems ⚫ Height of a projected object ⚫ Modeling with Quadratic Equations
- 3. Geometry Problems ⚫ Problems involving the area or volume of a geometric object or problems involving the Pythagorean Theorem often lead to quadratic equations. ⚫ As we did on applying linear equations, it’s important to ⚫ Read the problem. Make a diagram if necessary, so you understand what you are trying to find. ⚫ Assign a variable – remember, you can often express one quantity in terms of another. ⚫ Write an equation and solve it. ⚫ State your answer – are you answering the question? ⚫ Check – does your answer make sense?
- 4. Geometry Problems (cont.) Example: A piece of machinery is capable of producing rectangular sheets of metal such that the length is three times the width. Furthermore, equal-sized squares measuring 5 in. on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 1435 in3, what should the dimensions of the original piece of metal be?
- 5. Geometry Problems (cont.) ⚫ Read the problem: We need to find the dimensions of the original piece of metal. Draw a rectangle. ⚫ Assign a variable: Let x be the width; this means that the length would be represented as 3x. Each corner has a 5×5 square cut out. We then subtract that from the length and the width. 3x x 5 5 3x – 10 x – 10
- 6. Geometry Problems (cont.) ⚫ Write an equation. The volume of a rectangular solid is V = lwh. This means our equation is: 5 3x – 10 x – 10 ( )( )( )1435 3 10 10 5x x= − − 2 1435 15 200 500x x= − + 2 15 200 935 0x x− − = 2 3 40 187 0x x− − = ( )( ) ( ) 2 40 40 4 3 187 40 3844 40 62 2 3 6 6 x − − = = = 11 or 17 3 x = − X
- 7. Geometry Problems (cont.) ⚫ State the answer. The width is 17 in., so the length is 3(17) = 51 in. ⚫ Check: ( )( )( ) ( )( )( ) 3 51 10 17 10 5 41 7 5 1435 in V lwh= = − − = =
- 8. Pythagorean Theorem ⚫ The Pythagorean Theorem sets up the relationship between the sides of a right triangle. It states that ⚫ The hypotenuse will always be the longest side. In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the hypotenuse. hypotenuse (c) right angle a b 2 2 2 a b c+ =
- 9. Pythagorean Theorem (cont.) ⚫ Example: Erik finds a piece of property in the shape of a right triangle. He finds that the longer leg is 20 m longer than twice the length of the shorter leg. The hypotenuse is 10 m longer than the length of the longer leg. Find the lengths of the sides of the triangular lot. x 2x + 20 2x + 30 ( ) ( ) 2 22 2 20 2 30x x x+ + = +
- 10. Pythagorean Theorem (cont.) ⚫ The lengths of the sides of the property are 50 m, 120 m, and 130 m. ( ) ( ) 2 22 2 20 2 30x x x+ + = + 2 2 2 4 80 400 4 120 900x x x x x+ + + = + + 2 40 500 0x x− − = –500 –50 10 –40 ( )( )50 10 0x x− + = 50, 10x = −X ( )2 50 20 120 m+ = ( )2 50 30 130 m+ =
- 11. Height of a Projected Object ⚫ If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height of s0 feet, with initial velocity v0 feet per second is where t is the number of seconds after the object is projected. ⚫ The coefficient of t2, –16, is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon and other planets. 2 0 016 ,s t v t s= − + +
- 12. Height of a Projected Object ⚫ Example: If a projectile is shot vertically upward from the ground with an initial velocity of 100 ft per sec, neglecting air resistance, its height s (in feet) above the ground t seconds after projection is given by a) After how many seconds will it be 50 ft above the ground? b) How long will it take for the projectile to return to the ground? 2 16 100 .s t t= − +
- 13. Height of a Projected Object a) We must find value(s) of t so that height s is 50 ft. 2 50 16 100t t= − + 2 16 100 50 0t t− + = 2 8 50 25 0t t− + = ( ) ( ) ( )( ) ( ) 2 50 50 4 8 25 2 8 t − − − − = 50 1700 16 t = .55 or 5.70t t Both values are acceptable, since the projectile reaches 50 ft on the way up and the way down.
- 14. Height of a Projected Object b) When the projectile returns to the ground, its height will be 0 ft, so let s = 0 in the given equation. t = 0 represents when the projectile is launched, and 6.25 sec is when it returns to the ground. 2 0 16 100t t= − + 2 0 4 25t t= − ( )0 4 25t t= − 25 0 or 6.25 4 t t= = =
- 15. Height of a Project Object
- 16. Classwork ⚫ 10/14 Attendance Check/Progress Check ⚫ 1.5 Assignment (College Algebra) ⚫ Page 127: 6-18 (even); page 119: 32-56 (×4; omit 44); page 110: 70-90 (even; omit 82) ⚫ 1.5 Classwork Check ⚫ Quiz 1.4