1. In a circle, chords AB and CD intersect at point E.
If AE = x + 1, EB = x, CE = 2, and ED = 3, find
the value of x.
How would you solve this?

x = 2

Step-by-step explanation:

Given 2 intersecting chords, then the product of the parts of one chord is equal to the product of the parts of the other chord, that is

EB × AE = ED × CE

x(x + 1) = 3 × 2 = 6

x² + x = 6 ( subtract 6 from both sides )

x² + x - 6 = 0 ← in standard form

(x + 3)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 2 = 0 ⇒ x = 2

But x > 0 , then x = 2

whats the question?

step-by-step explanation:

the answer is 21%.