The length of a rectangle is 5 yd less than double the width, and the area of the rectangle is 52 yd^2 . How do you find the dimensions of the rectangle?

Define the variables first.

We could use two different variables, but we have been told how the length and width are related.

Let the width be #x" width is the smaller side"#
The length = #2x -5#

"Area = l x w" and the area is given to be 52 squ yards.

#A = x(2x-5) = 52#

#2x^2 -5x = 52" quadratic equation"#

#2x^2 -5x -52 = 0#

To factorise, find factors of 2 and 52 which cross-multiply and subtract to give 5.

#color(white)(xxx)(2)" "(52)#
#color(white)(xx.x) 2" 13 "rArr 1xx13 = 13#
#color(white)(xx.x) 1" 4 "rArr2xx4 = 8" "13-8 = 5#

We have the correct factors, now fill in the signs. We need -5.

#color(white)(xxx)(2)" "(-52)#
#color(white)(xx.x) 2" - 13 "rArr 1xx-13 = -13#
#color(white)(xx.x) 1" +4 "rArr2xx+4 = +8" "-13+8 = -5#

#(2x-13)(x+4) = 0#

Each factor could be equal to 0

#x = 6.5 or x = -4 # (reject)

The width = 6.5 yards. Now find the length: 6.5 x 2 -5 = 8 yards

Check:
Width = 6.5yds, length = 8yds
Area = 6.5 x 8 = 52

Let width be #=x#
Therefore, length #=2x -5#

We know that
#"Area" = "Length" xx "Width"#
Inserting given and assumed numbers we get

#52=(2x-5)xx x#
rearranging we obtain

#2x^2 -5x -52 = 0#

To factorize we use split the middle term method. We have two parts of middle term as #-13x and 8x#. The equation becomes

#2x^2-13x+8x-52 = 0#
Paring and taking out common factors we have
#x(2x-13)+4(2x-13) = 0#
#=>(2x-13)(x+4) = 0#

Setting each factor equal to #0#, we have two roots
#(2x-13)=0and (x+4) = 0#
#x = 13/2=6.5#
# x = -4 #, rejected as width can not be a #-ve# value

#:.#Width #= 6.5 yd#. And length#= 2xx6.5 -5 = 8 yd#

Check:
Area #= 8xx 6.5 = 52yd^2#