In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, BE=2x2−x , and DE=x2+6 .

2 answers:

8 0

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

4vir4ik [10]2 years ago

7 0

Answer:

The answer is 30, I took the test

Step-by-step explanation:

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