Answer:
[tex]Width=15\ inLength=20\ in[/tex]
Step-by-step explanation:
x^2=+5X=300
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 1
y=x+5 ___equation 2
substitute equation 2 in equation 1
[tex]A=x(x+5)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\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
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 \pm\sqrt{5^{2}-4(1)(-300)}} {2(1)}[/tex]
[tex]x=\frac{-5 \pm \sqrt{1225}} {2}[/tex]
[tex]x=\frac{-5 \pm 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
y=x+5
[tex]y=15+5=20\ in[/tex]
[tex]Width=15\ inLength=20\ in[/tex]
Answer:
Width = 15 inches
length = 20 inches
Step-by-step explanation:
Let
width = x
According to the question the length of the rectangle is 5 inches more than it width. Therefore,
length = x + 5
Area of a rectangle = LW
where
L = length
W = width
Area of the rectangle
x² + 5x = 300
x² + 5x - 300 = 0
Find the number you can multiply to give you -300 and add to give you 5. The numbers are -15 and 20
x² - 15x + 20x -300 = 0
x(x - 15) + 20(x - 15)
(x - 15)(x + 20) = 0
x = 15 or - 20
we can't use - 20 since it is negative.
Recall
width = x = 15
length = x + 5 = 15 + 5 = 20
Width = 15 inches
length = 20 inches
