Question 463216


If the side length of a square can be represented by {{{2x + 10}}} and its area is {{{256units^2}}}, find the value of {{{x}}}.

the area is {{{(2x + 10)(2x + 10)=256units^2}}}

{{{4x^2+20x+20x + 100=256units^2}}}

{{{4x^2+40x + 100-256units^2=0}}}


{{{4x^2+40x-156units^2=0}}}...use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-40 +- sqrt( 40^2-4*4*(-156) ))/(2*4) }}} 

{{{x = (-40 +- sqrt( 1600+2496))/8 }}} 

{{{x = (-40 +- sqrt( 4096))/8 }}} 

{{{x = (-40 +- 64)/8 }}} 

use only positive solution because the length cannot be negative


{{{x = (-40 + 64)/8 }}}

{{{x = 24/8 }}}

{{{x = 3 }}}



so, the length is: {{{2x + 10}}} ....->...{{{2*3 + 10=6+10=16units}}} 

the area is{{{16units*16units=256units^2}}}