Question

In parallelogram ABCD, diagonals AC and BD intersect at point E, BE =2x^2-3x, and DE=x^2+10. What is BD? Enter your answer in the box

Answer 1

Answer:

70

Step-by-step explanation:

In parallelogram ABCD, diagonals AC and BD intersect at point E

E is the midpoint of BD

So BE = DE

Given: BE =2x^2-3x, and DE=x^2+10

BE = DE

$2x^2-3x = x^2+10$ (solve for x)

Subtract x^2 and 10 on both sides

$2x^2-3x- x^2-10=0$

$x^2-3x-10=0$

Factor x^2-3x-10

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

x-5 =0  so x=5

x+2=0 so x=-2

Length x cannot be negative so we ignore x=-2

Lets plug in 5 for x  and find out BE  and DE

BE =$2x^2-3x= 2(5)^2-3(5)= 50-15= 35$

DE =$x^2+10= (5)^2+10=35$

BD = BE + DE= 35+35 = 70

So BD = 70