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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

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4.8 Polynomial Word Problems

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  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, …

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