Question

using the secant and segments theorem. what is the value of x? x = 2 x = 3 x = 4 x = 6

Explanation:

Step1: Apply secant - secant rule

If two secants are drawn to a circle from an exterior point, then $(EA)\times(ED)=(CA)\times(CB)$. Here, $EA = 14$, $ED=14$, $CA=21 + x$, and $CB=x$. So, $14\times14=(x + 21)\times x$.

Step2: Expand the equation

$196=x^{2}+21x$.

Step3: Rearrange to quadratic form

$x^{2}+21x - 196 = 0$.

Step4: Factor the quadratic equation

We need to find two numbers that multiply to - 196 and add up to 21. The numbers are 28 and - 7. So, $x^{2}+21x - 196=(x + 28)(x - 7)=0$.

Step5: Solve for x

Setting each factor equal to zero gives $x+28 = 0$ or $x - 7=0$. So, $x=-28$ or $x = 7$. Since length cannot be negative, we discard $x=-28$.

Answer:

$x = 7$ (Note: There seems to be an error in the provided multiple - choice options as 7 is not among them)