SOLUTION: a rectangle has a perimeter of 6x^3+9x^2-10x+5 and a length of x. Find the width of the rectangle when the length is 21 inches.

Algebra ->  Rectangles -> SOLUTION: a rectangle has a perimeter of 6x^3+9x^2-10x+5 and a length of x. Find the width of the rectangle when the length is 21 inches.      Log On


   

Question 1055024: a rectangle has a perimeter of 6x^3+9x^2-10x+5 and a length of x. Find the width of the rectangle when the length is 21 inches.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
2x%2B2%2A21=6x%5E3%2B9x%5E2-10x%2B5

6x%5E3%2B9x%5E2-10x-2x%2B5-42=0
6x%5E3%2B9x%5E2-12x-37=0
Can you find the zeros for this and pick one that will work?

y=6x%5E3%2B9x%5E2-12x-37
graph%28400%2C400%2C-2%2C30%2C-2%2C30%2C6x%5E3%2B9x%5E2-12x-37%29
graph%28400%2C400%2C-15%2C15%2C-15%2C15%2C6x%5E3%2B9x%5E2-12x-37%29

system%28x=1.727%2C+approximately%29
Put the function into google.com and see a better graph.

Answer by ikleyn(52793) About Me  (Show Source):
You can put this solution on YOUR website!
.
a rectangle has a perimeter of 6x^3+9x^2-10x+5 and a length of x. Find the width of the rectangle when the length is 21 inches.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

According to the condition, the perimeter of the rectangle is 

P = 6*21^3 + 9*21^2 - 10*21 + 5 = 59330 in.

Distract the doubled length (2*21 in) from the perimeter and divide the result by 2, and you will get the width.


The solution by "josgarithmetic" is incorrect.

He simply didn't give the labor to himself to read and to understand the condition.