MATH SOLVE

2 months ago

Q:
# The area of a rectangle is 77 m2 , and the length of the rectangle is 3 m more than twice the width. find the dimensions of the rectangle

Accepted Solution

A:

Answer:

Step-by-step explanation:Let x = width in ft y= length in ft Given, y = 3+2x ----(1)Area = xy = 77 ft2Using (1) to replace y3x+2x2=772x2+3x-77=0Solve this quadratic equation, a=2, b=3, c=-77x=(-3(+or-)sqrt(32-4*2*(-77)))/(2*2) = (-3 (+ or -) sqrt(9+616))/4 = (-3 (+ or -) sqrt(625))/4 = (-3 (+ or -) 25)/4x = (-3+25)/4 or x = (-3-25)/4 = 22/4 = -28/4x = 5.5 x = -7 x is the length and can't be negative. so, proceed with x= 5.5 xy = 77 y = 77/5.5 = 14 Dimensions of the rectangle are 5.5ft and 14 ft(Check: 2* 5.5 + 3 = 11+3 = 14)

Step-by-step explanation:Let x = width in ft y= length in ft Given, y = 3+2x ----(1)Area = xy = 77 ft2Using (1) to replace y3x+2x2=772x2+3x-77=0Solve this quadratic equation, a=2, b=3, c=-77x=(-3(+or-)sqrt(32-4*2*(-77)))/(2*2) = (-3 (+ or -) sqrt(9+616))/4 = (-3 (+ or -) sqrt(625))/4 = (-3 (+ or -) 25)/4x = (-3+25)/4 or x = (-3-25)/4 = 22/4 = -28/4x = 5.5 x = -7 x is the length and can't be negative. so, proceed with x= 5.5 xy = 77 y = 77/5.5 = 14 Dimensions of the rectangle are 5.5ft and 14 ft(Check: 2* 5.5 + 3 = 11+3 = 14)