Question

Sum of the areas of two squares is 468 m²
. If the difference of their perimeter is 24 m,
formulate the quadratic equation to find the sides of the two squares.​

Answer 1

Answer:

Step-by-step explanation:

let the side of of first square be x and of the second one be y

APQ, x^2+y^2=468..............( equation 1)       (area of square =side.side)

Also,4x-4y=24                                                (perimeter of square =4.side)

      4x-4y=24........(divide this equation by 4)

       x - y =6

       x=y+6

now substitute x in equation  1      

       (y+6)^2 +y^2=468

       y^2+24y+y^2=468

       2y^2+24y-468=0.............(divide this equation by 2)

       y^2+12y-234=0

       On solving by quadratic formula or splitting middle term , you will get the values as

x=18m

y=12m

hope this helps you!  

Answer 2

${x}^{2} + {y}^{2} = 468 - - - 1 \\ \\ 4x - 4y = 24 - - - 2 \\ \\ 4(x - y) = 24 \\ \\ x - y = 6 \\ \\ x = 6 + y - - - - 3 \\ \\ substitute \: eq - 3 \: in \: eq - 1 \\ \\ {(6 + y)}^{2} + {y}^{2} = 468 \\ \\ 36 + {y}^{2} + 12y + {y}^{2} = 468 \\ \\ 2 {y}^{2} + 12y = 432 \\ \\ common \: 2 \\ \\ {y}^{2} + 6y - 216 = 0 \\ \\ factorise \\ \\ {y}^{2} + 18y - 12y - 216 \\ \\ y(y + 18) - 12(y + 18) \\ \\ (y - 12)(y + 18) \\ \\ y = 12 \: or \: y = - 18$