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Quadratic Equation Find the length and width of a rectangle where the length is...
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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` .
Posted by sciencesolve on October 30, 2012 at 5:56 AM (Answer #1)
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.
Posted by eagudelo on October 30, 2012 at 2:19 AM (Answer #2)
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