Question

The length of a rectangle is 5 inches more than its vedth. The area of the rectangle is equal to

Answer 1

Answer:

• The width = x = 14

• The length = x+5 = 14+5 = 19

Step-by-step explanation:

Let  l  = length,  w = width,  A = area =  l×w , and  P  = perimeter =  2(l+w)

Let 'x' be the width

As the length 'l' of a rectangle is 5 inches more than its width.

so the length will be = x+5

As the Area of the rectangle is equal to  2 inches more than 4 times the perimeter.

A = 4P + 2

so the equation becomes

l × w = 4×2(l+w)+2

substituting w=x, l = x+5,

(x+5)x = 8(x+5+x)+2

x²+5x = 8(5+2x)+2

x²+5x = 40+16x+2

x²+5x = 16x+42

x²+5x-16x-42 =0

x²-11x-42=0

$x^2-11x=42$

$\mathrm{Add\:}a^2=\left(-\frac{11}{2}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-11x+\left(-\frac{11}{2}\right)^2=42+\left(-\frac{11}{2}\right)^2$

$x^2-11x+\left(-\frac{11}{2}\right)^2=\frac{289}{4}$

$\left(x-\frac{11}{2}\right)^2=\frac{289}{4}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{11}{2}=\sqrt{\frac{289}{4}}$

$x-\frac{11}{2}=\frac{\sqrt{289}}{\sqrt{4}}$

$x-\frac{11}{2}=\frac{\sqrt{289}}{2}$

$x-\frac{11}{2}=\frac{17}{2}$

$x-\frac{11}{2}+\frac{11}{2}=\frac{17}{2}+\frac{11}{2}$

$x=14$

similarly solving

$x-\frac{11}{2}=-\sqrt{\frac{289}{4}}$

$x-\frac{11}{2}=-\frac{17}{2}$

$x-\frac{11}{2}+\frac{11}{2}=-\frac{17}{2}+\frac{11}{2}$

$x=-3$

so

x = 14, or x = -3

As the width 'x' can not be negative.

so x = 14

Thus,

• The width = x = 14

• The length = x+5 = 14+5 = 19