Respuesta :
[tex]\bf \textit{area of a rectangle}\\\\
A=length\cdot width\qquad \boxed{A=65~in^2}\qquad 65=(\stackrel{length}{x+6})(\stackrel{width}{x-2})
\\\\\\
65=x^2+4x-12\implies 0=x^2+4x-77
\\\\\\
0=(x+7)(x-11)\implies x=
\begin{cases}
-7\\
\boxed{11}
\end{cases}[/tex]
it cannot be -7, because that would give a negative value for either dimension, and a dimension for the rectangle cannot be negative.
so, the dimensions would be (11) + 6 and (11) - 2.
it cannot be -7, because that would give a negative value for either dimension, and a dimension for the rectangle cannot be negative.
so, the dimensions would be (11) + 6 and (11) - 2.
Answer:
The dimensions of the rectangle is 13 length and 5 breadth.
Step-by-step explanation:
Given : The dimensions of a rectangle can be expressed as x+6 and x-2. If the area of the rectangle is 65 in².
To find : The dimensions of the rectangle ?
Solution :
Let length l= x+6
Breadth b= x-2
The area of the rectangle is A= 65 in².
The area of the rectangle is given by,
[tex]A=l\times b[/tex]
[tex]65=(x+6)\times (x-2)[/tex]
[tex]65=x^2-2x+6x-12[/tex]
[tex]x^2+4x-12-65=0[/tex]
[tex]x^2+4x-77=0[/tex]
Applying middle term split,
[tex]x^2+11x-7x-77=0[/tex]
[tex]x(x+11)-7(x+11)=0[/tex]
[tex](x+11)(x-7)=0[/tex]
[tex]x=-11,7[/tex]
The value of x=7 as we reject -11.
Length l= x+6=7+6=13 inches
Breadth b= x-2=7-2=5 inches
The dimensions of the rectangle is 13 length and 5 breadth.
