SOLUTION: The area of a rectangular field is 428 square meters and the width is 10m less than the length. What is the size of the width?

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: The area of a rectangular field is 428 square meters and the width is 10m less than the length. What is the size of the width?       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 953749: The area of a rectangular field is 428 square meters and the width is 10m less than the length. What is the size of the width?
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
L=length; W=width=L-10m; A=area=L*W
A=L%2AW
428m%5E2+=L%2AW Substitute for W
428m%5E2=L%2A%28L-10m%29
428m%5E2=L%5E2-10L Subtract 428m^2 from each side
0=L%5E2-10L-428m%5E2
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aL%5E2%2BbL%2Bc=0 (in our case 1L%5E2%2B-10L%2B-428+=+0) has the following solutons:

L%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-10%29%5E2-4%2A1%2A-428=1812.

Discriminant d=1812 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--10%2B-sqrt%28+1812+%29%29%2F2%5Ca.

L%5B1%5D+=+%28-%28-10%29%2Bsqrt%28+1812+%29%29%2F2%5C1+=+26.2837966537928
L%5B2%5D+=+%28-%28-10%29-sqrt%28+1812+%29%29%2F2%5C1+=+-16.2837966537928

Quadratic expression 1L%5E2%2B-10L%2B-428 can be factored:
1L%5E2%2B-10L%2B-428+=+1%28L-26.2837966537928%29%2A%28L--16.2837966537928%29
Again, the answer is: 26.2837966537928, -16.2837966537928. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-10%2Ax%2B-428+%29

L=26.28 The length is 26.28 meters.
W=L-10m=26.28m-10m=16.28m ANSWER: The width is 16.28 meters.
CHECK:
L%2AW=A
26.28m%2816.28m%29=428m%5E2
427.84m=428m Discrepancy from rounding.