Answer
Given the statement: In parallelogram ABCD, diagonals AC and BD intersects at point E.
Given that ABCD is a parallelogram.
As, we know that in a parallelogram diagonals bisect each other.
Since AC and BD intersect at E, and we get E is the mid point of both diagonals AC and BD.
⇒ BE = DE .....[1]
Substitute the given values of [tex]BE = 2x^2 -3x[/tex] and [tex]DE = x^2+10[/tex] in [1] we have;
[tex]2x^2-3x = x^2+10[/tex]
Subtract [tex]x^2[/tex] on both sides we get;
[tex]2x^2-3x-x^2 = x^2+10-x^2[/tex]
Simplify:
[tex]x^2-3x =10[/tex]
Subtract 10 on both sides we get;
[tex]x^2-3x-10 =0[/tex]
or
[tex]x^2-5x+2x-10 =0[/tex]
[tex]x(x-5)+2(x-5)=0[/tex]
[tex](x-5)(x+2)=0[/tex]
equate these factors equal to zero we get;
(x-5) = 0 and (x+2) = 0
we have;
x = 5 and x = -2
Since, x cannot be negative.
So, x =5
[tex]BE = DE = x^2+10 = (5)^2+10 = 25+10 = 35 units[/tex]
Diagonals BD = BE + DE = 35 + 35 =70 units.
Therefore, the value of BD = 70 units.
