Question

find the dimensions of a rectangle that has a length 3 inches longer than it’s width and has an area of 40in^2

Answer 1

Answer

length: 8 inch and width: 5 inch

Explanation

Let width be x,

then the length: x + 3

solve:

$\sf (x+3)(x) = 40$

$\sf x^2+3x = 40$

$\sf x^2+3x-40=0$

$\sf x^2+8x-5x-40=0$

$\sf x(x+8)-5(x+8)=0$

$\sf (x-5)(x+8)=0$

$\sf x = 5 \ or \ -8$

• As width cannot be negative, width will be 5.

Find for length:

$\sf x + 3$

$\sf 5+ 3$

$\sf 8$