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.

### 2 Answers | Add Yours

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` .**

*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=*

234=(18)(13)

*lw*234=234

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

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes