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sum of the areas of two squares is 468msquare. if the diff of there perimeter is 24m . formulate the quad eq to find the sides of the squares? plz.................. help
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x^2 + y^2 = 468 and 4x + 4y = 24 can you solve now?
\[x^2 + y^2 = 468\] \[4x + 4y = 24 \]
can you solve now?
oh ya m sry, it says difference, right!1 it must be a minus sign in between thats the difference of perimeters ;)
ha... then x^2 +y^2= 468 4(x+y)=24 x+y=6 squaring on both sides x^2 + y^2 +2xy= 36 2xy= 36 - (468) 2xy= -432 xy=-216 x(6-x)=-216 6x-x^2=-216 x^2-6x-216=0 solve it for x and put it in the above equation to get y.... the values may differ with actual but the process is correct i hope...
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