From 6ccb4b79aa1e290ade15c4e1b876302250471623 Mon Sep 17 00:00:00 2001
From: tkoz0 <13438573+tkoz0@users.noreply.github.com>
Date: Sat, 14 Dec 2024 01:00:57 -0500
Subject: [PATCH] roots

---
 academic/misc/2024-math-calendar/12-dec.html | 15 +++++++++++++++
 1 file changed, 15 insertions(+)

diff --git a/academic/misc/2024-math-calendar/12-dec.html b/academic/misc/2024-math-calendar/12-dec.html
index 79ad3f8..1ce9b4c 100644
--- a/academic/misc/2024-math-calendar/12-dec.html
+++ b/academic/misc/2024-math-calendar/12-dec.html
@@ -494,6 +494,21 @@ <h4>4. Combining results</h4>
 
 <h3 id="dec13">Dec 13</h3>
 
+\[\sqrt{x+\sqrt{16x-64}}+\sqrt{x-\sqrt{16x-64}}=6\]
+\[\left(x+\sqrt{16x-64}\right)+\left(x-\sqrt{16x-64}\right)
++2\sqrt{\left(x+\sqrt{16x-64}\right)\left(x-\sqrt{16x-64}\right)}=36\]
+\[2x+2\sqrt{x^2-(16x-64)}=36\]
+\[x+\sqrt{x^2-16x+64}=18\]
+\[x+\sqrt{(x-8)^2}=18\]
+\[x+|x-8|=18\]
+
+<p>
+If \(x-8&lt;0\) then the equation because \(x-(x-8)=18\Rightarrow8=18\) so no
+solution. If \(x-8\geq0\) then we find the solution
+\[x+(x-8)=18\Rightarrow2x-8=18\Rightarrow2x=26\Rightarrow x=13\]
+This can be checked to ensure it is not an extraneous solution.
+</p>
+
 <h3 id="dec14">Dec 14</h3>
 
 <h3 id="dec15">Dec 15</h3>