Answer:
[tex]Width=15\ in[/tex]
[tex]Length=20\ in[/tex]
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
[tex]A=xy[/tex] -----> equation A
[tex]y=x+5[/tex] ----> equation B
substitute equation B in equation A
[tex]A=x(x+5)[/tex]
[tex]A=x^{2}+5x[/tex]
[tex]x^{2} +5x=300[/tex] ------> given problem
so
the area of the rectangle is equal to
[tex]A=300\ in^{2}[/tex]
Solve the quadratic equation
[tex]x^{2} +5x-300=0[/tex]
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} +5x-300=0[/tex]
so
[tex]a=1\\b=5\\c=-300[/tex]
substitute in the formula
[tex]x=\frac{-5(+/-)\sqrt{5^{2}-4(1)(-300)}} {2(1)}[/tex]
[tex]x=\frac{-5(+/-)\sqrt{1225}} {2}[/tex]
[tex]x=\frac{-5(+/-)35} {2}[/tex]
[tex]x=\frac{-5+35} {2}=15[/tex]
[tex]x=\frac{-5-35} {2}=-20[/tex]
the solution is
[tex]x=15\ in[/tex]
Find the value of y
[tex]y=x+5[/tex]
[tex]y=15+5=20\ in[/tex]
