Respuesta :
Answer:
x-2 and x+10 side lengths should be used to model a rectangle.
Step-by-step explanation:
x-2 and x+10 side lengths should be used to model a rectangle.
(x + 6) and (x – 2) lengths should be used to model the rectangle.
Word problem leading to a Quadratic equation, using the Factorization method to solve the problem.
x^2+4x−12 = 0 (Quadratic equation)
Find the two numbers whose sum is 4, and product is -12 (The two numbers are 6 and -2)
x^2+6x - 2x −12 = 0
(x^2+6x) - (2x −12) = 0
x(x+6) - 2(x+ 6) = 0
Either (x-2) or (x+ 6) = 0
If (x-2) = 0, it implies x = 2
If (x+ 6) = 0, it implies x = -6
x = 2 or -6
Or
The negative sign in the second dimension (that is, -6) has to be ignored because it renders the answer invalid. Therefore, it becomes +6 since one of the dimensions of a rectangle cannot be a negative number.
Length = 6 sq. unit ; breadth = 2 sq. unit
This has got to be a misstatement of some other problem. You’re probably being asked to factor x2+4x−12. But the question as posted is nonsense. Pick any (non-negative) rectangle dimensions you like and there is a value for x (actually, two values) that yields that area. Even some “negative area” rectangles will have corresponding values for x .
For real values of x, the function
f(x)=x2+4x−12
is a parabola with a minimum value of −16 when x=−2 . When x≠−2 , f(x) takes on larger values, without bound.
Since the area of a rectangle cannot be negative, the smallest allowed physically meaningful value for f(x) is 0. We can factor f(x) to see the roots:
x2+4x−12=(x+6)(x−2)
This tells us that f(x)=0 when x=−6 or x=2 . Whenever x<−6 or x>2, we have a positive area. Since you haven’t told us what (if any) physical meaning there is to x, it’s hard to know which of these ranges is relevant. But note that we still have no idea of the rectangle dimensions. (There are an infinite number of rectangles that have an area of, say, 0.5.)
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