Question

a rectangle has a width of $x - 1$ inches and a length of $3x$ inches. find the values of $x$ that will produce a rectangle that has an area greater than 18 square inches. (1 point)
all values of $x$ such that $x > 3$
all values of $x$ such that $x > 2$
all values of $x$ such that $x < - 3$ or $x > 2$
all values of $x$ such that $x < - 2$ or $x > 3$

Explanation:

Step1: Find the area formula

The area \(A\) of a rectangle is \(A = \text{length}\times\text{width}\). Given width \(w=x - 1\) and length \(l = 3x\), so \(A=3x(x - 1)=3x^{2}-3x\).

Step2: Set up the inequality

We want \(A>18\), so \(3x^{2}-3x>18\). Divide both sides by \(3\) to get \(x^{2}-x - 6>0\).

Step3: Factor the quadratic

Factor \(x^{2}-x - 6=(x - 3)(x + 2)\). So the inequality is \((x - 3)(x + 2)>0\).

Step4: Find the critical points

Set \((x - 3)(x + 2)=0\), the critical points are \(x = 3\) and \(x=-2\).

Step5: Test intervals

• For \(x<-2\), let \(x=-3\), \((-3 - 3)(-3+2)=(-6)\times(-1) = 6>0\).
• For \(-2
• For \(x>3\), let \(x = 4\), \((4 - 3)(4 + 2)=1\times6 = 6>0\).

Also, consider the domain of the rectangle: width \(x-1>0\) (since width cannot be negative) gives \(x>1\), and length \(3x>0\) gives \(x>0\). Combining with the inequality \((x - 3)(x + 2)>0\) and the domain of the rectangle (width and length must be positive, \(x-1>0\) implies \(x>1\)), we still get \(x>3\) (because when \(x>3\), both width \(x - 1>0\) and length \(3x>0\))

Answer:

all values of \(x\) such that \(x>3\)