Question 351137
Start with the formula for the area of a rectangle:
{{{A = L*W}}} Where L = length and W = width.
The area, A, is given as A = 4800 sq.cm.
The width, W, is given as:
{{{W = (1/2)L+20}}} "The width ... is 20cm more than half the length."
Let's express the above in terms of the length, L.
{{{W = (1/2)L+20}}} Multiply through by 2 to clear the fraction.
{{{2W = L+40}}} Subtract 40 from both sides.
{{{2W-40 = L}}} Now substitute this into the formula for L.
{{{A = L*W}}} Substitute A = 4800, L = 2W-40
{{{4800 = (2W-40)*W}}} Simplify.
{{{4800 = 2W^2-40W}}} Now subtract 4800 from both sides.
{{{2W^2-40W-4800 = 0}}} Divide by 2 to simplify a bit.
{{{W^2-20W-2400 = 0}}} Factor this quadratic equation.
{{{(W+40)(W-60) = 0}}} Apply the zero product rule.
{{{W+40 = 0}}} or {{{W-60 = 0}}} so...
{{{W = -40}}} or {{{W = 60}}} Discard the negative solution as width is a positive quantity.
The width is 60cm.
Note: This is certainly well beyond grade 4 level math in american schools.