Question 1146135

if the length and breadth of a rectangle are {{{(2x+5)cm}}} and {{{( 2x-1)cm}}} respectively, the area of rectangle is

{{{(2x+5)*( 2x-1)}}}

the area of square {{{(x+1)cm}}} is {{{(x+1)^2}}}

the area of rectangle is three times the area of square {{{(x+1)cm}}}, we have

{{{(2x+5)*( 2x-1)=3(x+1)^2}}}......simplify

{{{(2x+5)*( 2x-1)=3(x+1)^2}}}

{{{4x^2 + 8x - 5=3(x^2+2x+1)}}}

{{{4x^2 + 8x - 5=3x^2+6x+3}}}

{{{4x^2 + 8x - 5-3x^2-6x-3=0}}}


{{{x^2 + 2x - 8=0}}} => shows that is reduces to {{{x^2+2x-8=0}}}