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Quadratic Equation Find the length and width of a rectangle where the length is...

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paulbarton

Posted October 29, 2012 at 10:19 PM via web

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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.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 30, 2012 at 5:56 AM (Answer #1)

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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` .

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eagudelo | Student, College Freshman

Posted October 30, 2012 at 2:19 AM (Answer #2)

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l=x+8, w=x+3, A=234
A=lw
234=(x+8)(x+3)
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.
l=x+8
l=10+8=18
w=x+3
w=10+3=13

To prove that the length and width are right, plug them in the area equation.
A=lw
A=234
234=lw
234=(18)(13)
234=234

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

 

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