4.8 Polynomial Word Problems

a) Define the variable, b) Write the equation, and c) Solve the problem. 1) 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 x = –7 x = 6 x2x2 Check! (–7) = = 42 (6) = = 42 c) –7 or 6 b)

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

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 x 2 + y 2 = 490 x 2 + (28 – x) 2 = 490 x – 56x + x 2 = 490 2x 2 – 56x = 0 x 2 – 28x = 0 (x – 7)(x – 21) =0 x – 7 = 0 x – 21 = 0 x = 7 or x = 21 Check! The numbers are c) 7 & 21 b)  y = 28 – x ( )

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

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