Answer:
Step-by-step explanation:
We are given a parallelogram ABCD with diagonals AC and BD intersect at point E.
[tex]AE=x^2-16.[/tex] , and CE=6x .
Note: The diagonals of a parallelogram intersects at mid-point.
Therefore, AE = EC.
Plugging expressions for AE and EC, we get
[tex]x^2-16=6x.[/tex]
Subtracting 6x from both sides, we get
[tex]x^2-16-6x=6x-6x[/tex]
[tex]x^2-6x-16=0[/tex]
Factoriong quadratic by product sum rule.
We need to find the factors of -16 that add upto -6.
-16 has factors -8 and +2 that add upto -6.
Therefore, factor of [tex]x^2-6x-16=0[/tex] quadratic is (x-8)(x+2)=0
Setting each factor equal to 0 and solve for x.
x-8=0 => x=8
x+2=0 => x=-2.
We can't take x=-2 as it's a negative number.
Therefore, plugging x=8 in EC =6x, we get
EC = 6(8) = 48.
