Question

math1111: college algebra(11446)
lesson 14 - solving polynomial equations by factoring
lesson 14 - solving polynomial equations by factoring
score: 82/100 answered: 16/19
question 17
the length of a rectangle is 4 inches longer than it is wide. if the area is 165 square inches, what are the dimensions of the rectangle?
the width is
select an answer
the length is
select an answer
question help: video message instructor

Explanation:

Step1: Define variables

Let the width of the rectangle be $x$ inches. Then the length is $x + 4$ inches.

Step2: Set up the area - equation

The area of a rectangle is $A=\text{length}\times\text{width}$. So we have the equation $x(x + 4)=165$.

Step3: Expand the equation

Expand $x(x + 4)$ to get $x^{2}+4x=165$, and then rewrite it in standard quadratic - form $x^{2}+4x−165 = 0$.

Step4: Factor the quadratic equation

We need to find two numbers that multiply to $- 165$ and add up to $4$. The numbers are $15$ and $-11$ since $15\times(-11)=-165$ and $15+( - 11)=4$. So, $x^{2}+4x−165=(x + 15)(x - 11)=0$.

Step5: Solve for $x$

Set each factor equal to zero:
If $x+15 = 0$, then $x=-15$. But since the width cannot be negative, we discard this solution.
If $x - 11=0$, then $x = 11$.

Step6: Find the length

The length is $x + 4$. Substitute $x = 11$ into $x + 4$, we get the length $=11 + 4=15$.

Answer:

The width is 11 inches.
The length is 15 inches.