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.
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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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=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.