The length of a rectangle is 5 m more than its width. If the area of the rectangle is 15 m2, what are the dimensions of the rectangle, to the nearest tenth of a metre?

#color(blue)("Developing the equation")#

Let length be #L#
Let width be #w#

Let area be #a#

Then #a=Lxxw# ............................Equation(1)

But in the question it states:
"The length of a rectangle is 5m more than its width"#->L=w+5#

So by substituting for #L# in equation(1) we have:

#a=Lxxw" "->" "a=(w+5)xxw#

Written as: #a=w(w+5)#

We are told that #a=15m^2#

#=>15=w(w+5)....................Equation(1_a)#
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#color(blue)("Solving for value of width")#

Multiply out the bracket

#15=w^2+5w#

Subtract 15 from both sides

#w^2+5w-15=0#

Not that #3xx5=15# However, #3+-5!=5#

So using the standardised formula:
#y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=1; b=5; c=-15#

#=>x=(-5+-sqrt((5)^2-4(1)(-15)))/(2(1))#

#=>x=-5/2+-sqrt(85)/2#

A negative value is not logical so we use

#x=-5/2+sqrt(85)/2" "=" "2.109772..#

#color(green)("The question instructs that are to use the nearest 10th")#

#"width "= x=2.1" "# rounded to 1 decimal place
#color(red)(" "uarr)#
#color(red)(" This comment is very important")#
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#color(blue)("Solving for value of length")#

#a=Lxxw" "-> 15=Lxx2.109772..#

#=>L=15/2.109772.. = 7.1.9772..#

length# = 7.1 # rounded to 1 decimal place