Question

The length of a rectangle is 5 inches more than its width, x. The area of a rectangle can be represented by the equation x 2 + 5x = 300. What are the measures of the width and the length? Width = a0 inches Length = a1 inches NEXT QUESTION

Answer 1

Area=length*width
length=5+width
area=(5+w)(w)
area=5w+w^2
area=x^2+5x=300
x^2+5x-300=0
(x+20)(x-15)=0
x=15
w=15
l=5+15
l=20

Answer 2

Answer:

$Width=15\ in$

$Length=20\ in$

Step-by-step explanation:

Let

x-----> the width of the rectangle

y----> the length of the rectangle

we know that

the area of the rectangle is equal to

$A=xy$ -----> equation A

$y=x+5$ ----> equation B

substitute equation B in equation A

$A=x(x+5)$

$A=x^{2}+5x$

$x^{2} +5x=300$ ------> given problem

so

the area of the rectangle is equal to

$A=300\ in^{2}$

Solve the quadratic equation

$x^{2} +5x-300=0$

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +5x-300=0$

so

$a=1\\b=5\\c=-300$

substitute in the formula

$x=\frac{-5(+/-)\sqrt{5^{2}-4(1)(-300)}} {2(1)}$

$x=\frac{-5(+/-)\sqrt{1225}} {2}$

$x=\frac{-5(+/-)35} {2}$

$x=\frac{-5+35} {2}=15$

$x=\frac{-5-35} {2}=-20$

the solution is

$x=15\ in$

Find the value of y

$y=x+5$

$y=15+5=20\ in$