1 / 6

4.8 Polynomial Word Problems

4.8 Polynomial Word Problems. a) Define the variable, b) Write the equation, and c) Solve the problem. The sum of a number and its square is 42. Find the number. a) Let x = a number x + x 2 = 42 x 2 + x – 42 = 0 (x + 7)(x – 6) =0 x + 7 = 0 x – 6 = 0

rubaina-manaf
Télécharger la présentation

4.8 Polynomial Word Problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 4.8 Polynomial Word Problems

  2. a) Define the variable, b) Write the equation, and c) Solve the problem. • The sum of a number and its square is 42. Find the number. a) Let x = a number x + x2 = 42 x2 + x – 42 = 0 (x + 7)(x – 6) =0 x + 7 = 0 x – 6 = 0 x = –7 x = 6 x2 Check! (–7)2 + 7 = 49 + 7 = 42 (6)2 + 6 = 36 + 6 = 42 c) –7 or 6 b)

  3. 2) The length of a rectangle is 11 cm more than its width. The area of the rectangle is 2040 cm2. Find the dimensions of the rectangle. a) Let L = length & W = width L = W + 11 Area of rectangle = L • W L • W = 2040 (W + 11) • W = 2040 W2 + 11W – 2040 = 0 (W + 51)(W – 40) = 0 W + 51 = 0 W – 40 = 0 W = –51 W = 40 ( ) Dimensions can’t be (–) b) Plug in to find L: L = 40 + 11 = 51 40 cm x 51 cm

  4. 3) The sum of two numbers is 28. The sum of the squares of the two numbers is 490. Find the numbers. a) Let x = one number & y = other number x + y = 28 x2 + y2 = 490 x2 + (28 – x)2 = 490 x2 + 784 – 56x + x2 = 490 2x2 – 56x + 294 = 0 x2 – 28x + 147 = 0 (x – 7)(x – 21) =0 x – 7 = 0 x – 21 = 0 ( )  y = 28 – x b) x = 7 or x = 21 Check! The numbers are c) 7 & 21

  5. 4) The sum of the squares of two consecutive integers is 313. Find the integers. a) Let x = 1st integer & x + 1 = 2nd integer x2 + (x + 1)2 = 313 x2 + x2 + 2x + 1 = 313 2x2 + 2x – 312 = 0 x2 + x – 156 = 0 (x + 13)(x – 12) =0 x + 13 = 0 x – 12 = 0 x = –13 x = 12 b) When x = –13 x + 1 = –12 When x = 12 x + 1 = 13 The numbers are c) –13 & –12 or 12 & 13

  6. Homework #8 Polynomial Word Problems WS Remember Consecutive Odd Integers: x, x + 2, x + 4, …

More Related