In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, BE=2x2−3x , and DE=x2+10 .

What is BD ?

AngelRay
Nov 4, 2017

#2**+1 **

The diagonals of a parallelogram bisect each other....therefore...

2x^2 - 3x = x^2 + 10 simplify as

x^2 - 3x - 10 = 0 factor

(x - 5) ( x + 2) = 0

Setting each factor to 0 and solving for x, we have that x = 5 or x = -2

Therefore we have two possible values for BD

If x = 5 ...BD = 2 [ (5)^2 + 10] = 2 [ 35 ] = 70 units

If x = - 2, .....BD = 2 [ (-2)^2 + 10 ] = 2 [ 14 ] = 28 units

CPhill
Nov 5, 2017

#1**0 **

BD = \(\sqrt{3x^{2}-3x+10}\) . This is because, with the Pythagorean theorem, you add the 2 legs of a right triangle and take the square root to get the hypotenuse of it. This only applies if ABCD is a rhombus.

helperid1839321
Nov 4, 2017

#2**+1 **

Best Answer

The diagonals of a parallelogram bisect each other....therefore...

2x^2 - 3x = x^2 + 10 simplify as

x^2 - 3x - 10 = 0 factor

(x - 5) ( x + 2) = 0

Setting each factor to 0 and solving for x, we have that x = 5 or x = -2

Therefore we have two possible values for BD

If x = 5 ...BD = 2 [ (5)^2 + 10] = 2 [ 35 ] = 70 units

If x = - 2, .....BD = 2 [ (-2)^2 + 10 ] = 2 [ 14 ] = 28 units

CPhill
Nov 5, 2017