Question

Please please help me

Answer 1

Answer:

x = 5.5

Step-by-step explanation:

Given 2 secants intersecting the circle from a point outside the circle then

The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is

x(x + 14) = 6(6 + 12)

x² + 14x = 6 × 18 = 108 ( subtract 108 from both sides )

x² + 14x - 108 = 0 ← in standard form

with a = 1, b = 14 and c = - 108

Using the quadratic formula to solve for x

x = ( - b ± $\sqrt{b^2-4ac}$ ) / 2a

  = ( - 14 ± $\sqrt{14^2-(4(1)(-108)}$ ) / 2

  = ( - 14 ± $\sqrt{196+432}$ ) / 2

  = ( - 14 ± $\sqrt{628}$ ) / 2

  x = $\frac{-14-\sqrt{628} }{2}$ or x = $\frac{-14+\sqrt{628} }{2}$

  x = - 19.5 or x = 5.5 ( to 1 dec. place )

However x > 0 ⇒ x = 5.5