Given:
Two secants intersecting each other outside the side the circle.
To find:
The value of x.
Solution:
Using Intersecting Secant Theorem, we get
[tex]x(x+8)=4(4+8)[/tex]
[tex]x^2+8x=4(12)[/tex]
[tex]x^2+8x=48[/tex]
[tex]x^2+8x-48=0[/tex]
Using splitting the middle terms, we get
[tex]x^2+12x-4x-48=0[/tex]
[tex]x(x+12)-4(x+12)=0[/tex]
[tex](x+12)(x-4)=0[/tex]
Using zero product property, we get
[tex](x+12)=0\text{ and }(x-4)=0[/tex]
[tex]x=-12\text{ and }x=4[/tex]
Side length cannot be negative, i.e., [tex]x\neq -12[/tex].
Therefore, the only value of x is 4.
