In parallelogram abcd , diagonals ac¯¯¯¯¯ and bd¯¯¯¯¯ intersect at point e, be=2x2−x , and de=x2+6 . what is bd ?

The diagonals of a parallelogram bisect each other. So we just equate the two equations BE and DE 2x^2 - x = x^2 + 6 2x^2 - x - x^2 - 6 = 0 x^2 - x - 6 = 0 This is quadratic eqn. So we obtain. (x - 3) ( x + 2) = 0 Setting each to 0 and solving for x, we have that x = 3 and x = -2 So we have two possible values for BD. But Since BE = DE then we can simply double each If x = 3 BD = 2 (2(3)^2 - 3) = 2 ( 18 - 3) = 30 units If x = - 2 BD = 2((-2)^2 -3 ) = 2 (8 - 3) = 10 units

No a triangle cannot be formed because when you add the two smallest number together (which would be 11) the have to be bigger than the number it is "against" ( which would be 12)