Question 171781
My appologies to Nerdy Bill but the solution is a bit off.
From the description of the rectangle and its diagonals, the one diagonal is {{{RT = 3x^2}}} but the other diagonal QS is really given as {{{QC = 5x+4}}} and this is but half of the complete diagonal QS, so the equation becomes:
{{{RT = 2QC}}} or
{{{3x^2 = 2(5x+4)}}} Rewriting this it becomes:
{{{3x^2-10x-8 = 0}}} ...and this is factorable to:
{{{(3x+2)(x-4) = 0}}} and so...
{{{x = -2/3}}} or {{{x = 4}}} and, as you pointed out, you can discard the negative quantity as we are talking about lengths, so...
x = 4

Check:
{{{3x^2 = 3(4)^2}}} = {{{3(16) = 48}}}
{{{2(5x+4) = 2(20+4)}}} = {{{2(24) = 48}}}