Polishing

notebook/agentchat_MathChat.ipynb

@@ -136,7 +136,7 @@
    "source": [
     "### Example 1\n",
     "\n",
-    "Problem: Find all $x$ that satisfy the inequality $(2x+10)(x+3)<(3x+9)(x+8)$. Express your answer in interval notation.\n",
+    "Problem: Find all $x$ that satisfy the inequality `$(2x+10)(x+3)<(3x+9)(x+8)$`. Express your answer in interval notation.\n",
     "\n",
     "Correct Solution: \n",
     "We have \\begin{align*} (2x+10)(x+3)&<(3x+9)(x+8) \\quad \\Rightarrow\n",
@@ -145,7 +145,7 @@
     "\\\\ (2x+10-(3x+24))(x+3)&<0 \\quad \\Rightarrow\n",
     "\\\\ (-x-14)(x+3)&<0 \\quad \\Rightarrow\n",
     "\\\\ (x+14)(x+3)&>0.\n",
-    "\\end{align*} This inequality is satisfied if and only if $(x+14)$ and $(x+3)$ are either both positive or both negative.  Both factors are positive for $x>-3$ and both factors are negative for $x<-14$.  When $-14<x<-3$, one factor is positive and the other negative, so their product is negative.   Therefore, the range of $x$ that satisfies the inequality is $ \\boxed{(-\\infty, -14)\\cup(-3,\\infty)} $."
+    "\\end{align*} This inequality is satisfied if and only if $(x+14)$ and $(x+3)$ are either both positive or both negative.  Both factors are positive for $x>-3$ and both factors are negative for `$x<-14$`.  When `$-14<x<-3$`, one factor is positive and the other negative, so their product is negative.   Therefore, the range of $x$ that satisfies the inequality is $ \\boxed{(-\\infty, -14)\\cup(-3,\\infty)} $."
    ]
   },
   {
@@ -971,7 +971,7 @@
     "Problem: Find all numbers $a$ for which the graph of $y=x^2+a$ and the graph of $y=ax$ intersect. Express your answer in interval notation.\n",
     "\n",
     "\n",
-    "Correct Solution: If these two graphs intersect then the points of intersection occur when  \\[x^2+a=ax,\\] or  \\[x^2-ax+a=0.\\] This quadratic has solutions exactly when the discriminant is nonnegative: \\[(-a)^2-4\\cdot1\\cdot a\\geq0.\\] This simplifies to  \\[a(a-4)\\geq0.\\] This quadratic (in $a$) is nonnegative when $a$ and $a-4$ are either both $\\ge 0$ or both $\\le 0$. This is true for $a$ in $$(-\\infty,0]\\cup[4,\\infty).$$ Therefore the line and quadratic intersect exactly when $a$ is in $\\boxed{(-\\infty,0]\\cup[4,\\infty)}$.\n"
+    "Correct Solution: If these two graphs intersect then the points of intersection occur when  \\[x^2+a=ax,\\] or  \\[x^2-ax+a=0.\\] This quadratic has solutions exactly when the discriminant is nonnegative: \\[(-a)^2-4\\cdot1\\cdot a\\geq0.\\] This simplifies to  \\[a(a-4)\\geq0.\\] This quadratic (in $a$) is nonnegative when $a$ and $a-4$ are either both $\\ge 0$ or both $\\le 0$. This is true for $a$ in $$(-\\infty,0]\\cup[4,\\infty).$$ Therefore the line and quadratic intersect exactly when $a$ is in `$\\boxed{(-\\infty,0]\\cup[4,\\infty)}$`.\n"
    ]
   },
   {