Question 846781
the area of a rectangle is 12 square inches. length is 5 more than twice the width. find the length and the width

-------------------------------------------------------


Let x = width


"length is 5 more than twice the width" so 2x+5 is the length


{{{A = LW}}}


{{{12 = (2x+5)(x)}}}


{{{(2x+5)(x) = 12}}}


{{{x(2x+5) = 12}}}


{{{2x^2+5x = 12}}}


{{{2x^2+5x - 12 = 0}}}


{{{2x^2+8x - 3x - 12 = 0}}}


{{{(2x^2+8x) + (-3x - 12) = 0}}}


{{{2x(x+4) + (-3x - 12) = 0}}}


{{{2x(x+4) -3(x + 4) = 0}}}


{{{(2x-3)(x+4) = 0}}}


{{{2x-3=0}}} or {{{x+4 = 0}}}


{{{2x=3}}} or {{{x = -4}}}


{{{x=3/2}}} or {{{x = -4}}}


Toss the negative solution. A negative width makes no sense at all.


So the width is {{{3/2}}} inches, {{{1&1/2}}} inches or {{{1.5}}} inches


{{{L = 2x + 5}}}


{{{L = 2(3/2) + 5}}}


{{{L = 6/2 + 5}}}


{{{L = 3 + 5}}}


{{{L = 8}}}


and the length is 8 inches