Question

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.

Accepted Answer

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 .

Answer

l=x+8, w=x+3, A=234A=lw234=(x+8)(x+3)234=x^2 + 3x +8x + 24234=x^2 +11x +240=x^2 +11x - 210Use 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))/2x=(-11+or- 31)/2x=20/2 or x=-42/2x=10 or x=-21But 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+8l=10+8=18w=x+3w=10+3=13
To prove that the length and width are right, plug them in the area equation.A=lwA=234234=lw234=(18)(13)234=234
So the lenth is 18m and the width is 13m.

Math

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.

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