Difference between revisions of "Power of a Point Theorem/Introductory Problem 1"
Coltsfan10 (talk | contribs) (→Solution) |
Coltsfan10 (talk | contribs) (→Solution) |
||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
| − | Applying the [[Power of a Point Theorem]], we get <math> 3\cdot(3+5) = x (x+10) \rightarrow x^2 + 10x - 24 = 0 </math>. This factors as <math> (x+12)(x-2) = 0 </math>. We discard the negative solution since distance must be positive. Thus <math> \fbox{ | + | Applying the [[Power of a Point Theorem]], we get <math> 3\cdot(3+5) = x (x+10) \rightarrow x^2 + 10x - 24 = 0 </math>. This factors as <math> (x+12)(x-2) = 0 </math>. We discard the negative solution since distance must be positive. Thus <math> x=\fbox{2} </math>. |
''Back to the [[Power of a Point Theorem]].'' | ''Back to the [[Power of a Point Theorem]].'' | ||
Latest revision as of 14:15, 23 March 2020
Problem
Find the value of 
in the following diagram:
Solution
Applying the Power of a Point Theorem, we get 
. This factors as 
. We discard the negative solution since distance must be positive. Thus 
.
Back to the Power of a Point Theorem.
