Quadratic Equation Find the length and width of a rectangle where the length is represented by (x+8), the width is represented by (x+3) and the area is equal to 234m^2.  Please explain fully. 

Expert Answers info

sciencesolve eNotes educator | Certified Educator

calendarEducator since 2011

write5,349 answers

starTop subjects are Math, Science, and Business

You need to remember the formula of area of rectangle such that:

A = length*width

Since the problem provides the value of area of `234 m^2` , the length of `(x+8)`  and the width of `(x+3), ` you need to substitute these values in equation of area such that:

`234 = (x+8)(x+3)`

Opening the brackets yields:

`234 = x^2 + 3x + 8x + 24 => x^2 + 11x - 210 = 0`

You may use quadratic formula to find x such that:

`x_(1,2) = (-b+-sqrt(b^2-4ac))/(2a)`

You need to identify a,b,c such that:

`a=1 , b = 11 , c = -210`

`x_(1,2) = (-11+-sqrt(121+840))/2 => x_(1,2) = (-11+-sqrt961)/2`

`x_(1,2) = (-11+-31)/2 => x_1 = 10 , x_2 = -21`

Since the length and the width are `(x+8)`  and (x+3), then only the value x = 10 is valid.

`x+8 = 10+8 = 18`

`x+3 = 10+3 = 13`

Hence, evaluating the length and the width of the given rectangle, under the given conditions, yields `x+8=18 m`  and `x+3=13 m` .

check Approved by eNotes Editorial

eagudelo | Student

l=x+8, w=x+3, A=234
234=x^2 + 3x +8x + 24
234=x^2 +11x +24
0=x^2 +11x - 210
Use the quadratic formula x=(-b+or- sqrt(b^2 -4ac))/(2a)
so, x=(-11+or- sqrt(11^2 - 4(1)(-210)))/(2(1))
x=(-11+or- sqrt(961))/2
x=(-11+or- 31)/2
x=20/2 or x=-42/2
x=10 or x=-21
But the right value for x is 10 because you can't have a negative measurement (once you plug in x for l and w.)

Plug in x for l and w.

To prove that the length and width are right, plug them in the area equation.

So the lenth is 18m and the width is 13m.


check Approved by eNotes Editorial