Question

Please help me with this

Answer 1

This is a secant-tangent problem.

(Whole)(outside) = (tangent)^2

(x + 12)(x) = 8^2

x^2 + 12x = 64

x^2 + 12x - 64 = 0

In this quadratic equation, we see that a = 1, b = 12 and c = -64.

Plug those values into the quadratic formula and solve for x.

Answer 2

Answer:

x = 4

Step-by-step explanation:

Given a secant and a tangent drawn from an external point to the circle, then

The square of the measure of the tangent is equal to the product of the external part and the entire secant, that is

x(x + 12) = 8²

x² + 12x = 64 ( subtract 64 from both sides )

x² + 12x - 64 = 0 ← in standard form

with a = 1, b = 12 and c = - 64

Using the quadratic formula to solve for x

x = ( - 12 ± $\sqrt{12^2-(4(1)(-64)}$ ) / 2

  = ( - 12 ± $\sqrt{144+256}$ ) / 2

  = ( - 12 ± $\sqrt{400}$ ) / 2

x = $\frac{-12-20}{2}$ or x = $\frac{-12+20}{2}$

x = - 16 or x = 4

However x > 0 ⇒ x = 4