Find numbers that multiply to 28 and add them to see if they add to 8 28= 1 and 28=29 not 8 2 and 14=16 not 8 4 and 7=11 not 8 that's it' no 2 numbers we must use quadratic formula
x+y=8 xy=28
x+y=8 subtract x fromb oths ides y=8-x subsitute x(8-x)=28 distribute 8x-x^2=28 add x^2 to both sides 8x=28+x^2 subtract 8x x^2-8x+28=0
if you have ax^2+bx+c=0 then x= so if we have 1x^2-8+28=0 then a=1 b=-8 c=28
x= x= x= x= x= there are no real numbers that satisfy this
No it is not. If you had two fixed diagonals which are attached in the center, when they move it creates different parallelograms. The area would be close to zero if they were parallel.