in parallelogram abcd , diagonals ac⎯⎯⎯⎯⎯ and bd⎯⎯⎯⎯⎯ intersect at point e, be=x2−40 , and de=6x . what is bd ? enter your answer in the box. bd = units

  bd = 120 units

Step-by-step explanation:

The diagonals of a parallelogram bisect each other. This means ...

  be = de

  x^2 -40 = 6x . . . . . . substitute given expressions

  x^2 -6x -40 = 0 . . . . subtract 6x

  (x -10)(x +4) = 0 . . . . . factor

Solutions are the values of x that make the factors zero. These are -4 and +10. Only the positive value is of interest here, since the segment lengths must be positive.

  de = 6x = 6(10) = 60

The length of bd is twice this length, so ...

  bd = 120 units

The correct answer is :  AC = 108 units.

Step-by-step explanation:

We are given a parallelogram ABCD with diagonals AC and BD intersect at point E.

AE=x^2-40 , and CE=6x .

Note: The diagonals of a parallelogram intersects at mid-point.

Therefore, AE = EC.

Plugging expressions for AE and EC, we get

X^2-40=6x

Subtracting 6x from both sides, we get

X^2-40-6x=6x-6x

X^2-6x-40=0

Factoriong quadratic by product sum rule.

We need to find thej o  of -16 that add upto -6.

-16 has factors -10 and +(-4) that Adda upto -6.èb

Therefore, factor of x^2-6x-40n=0 quadratic is (x-10)(x+(-4))=0

Setting each factor equal to 0 and solve for x.

x-10=0  => x+(-4)=0

x+10=0  => x=-4.

We can't take x=-4 as it's a negative number.

Therefore, plugging x=10 in EC =6x, we get

EC = 6(10) = 60.

AC = AE + EC = 48+60 = 108 units.