From 754cf7230a5d9028cab033937d69a9e336c47832 Mon Sep 17 00:00:00 2001 From: Kenneth Odle Date: Sun, 16 Jul 2023 15:32:26 -0400 Subject: [PATCH] Typofixes --- complete-the-square.tex | 17 +++++++++-------- 1 file changed, 9 insertions(+), 8 deletions(-) diff --git a/complete-the-square.tex b/complete-the-square.tex index 977d21f..af60cc5 100644 --- a/complete-the-square.tex +++ b/complete-the-square.tex @@ -89,8 +89,8 @@ Because any number times zero is equal to zero, we can conclude that either $x-3 \begin{align}\label{eq3} \begin{split} -x-3 &0 \\ -x=3 +x-3 & =0 \\ +x &= 3 \end{split} \end{align} @@ -109,7 +109,7 @@ Using either of these values for $x$ in equation \ref{eq2} will result in one of For equations that are not easily factored, a general solution called ``the quadratic equation'' can be used to solve any quadratic. -For any equation of the form $ax^s+bx+c=0$, the solution can be found by using: +For any equation of the form $ax^2+bx+c=0$, the solution can be found by using: \begin{equation} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} @@ -129,25 +129,26 @@ Here is the general procedure: \item Divide $b$ by 2, square it, and add it to both sides of equation. \item Write the left side as a perfect square. \item Take the square root of both sides. +\item Solve for $x$. \end{enumerate} -\paragraph{Note:} Our goal is to make the \textit{left} side of the equation into a square. However, the right side will often not be a perfect square, as in \ref{eq1}. This means that we will have a $\sqrt{~~}$ sign in our solution. +\paragraph{Note:} Our goal is to make the \textit{left} side of the equation into a square. However, the right side will often not be a perfect square, as in equation set~(\ref{eq1}). This means that we will have a $\sqrt{~~}$ sign in our solution. \subsection{Example \#1} For our first example, we'll use an equation in which $a$ is already 1, so we can skip the first step. Our equation is \[x^2+4x+1=0\] -The second step is the move the constant term to the right side of the equation by subtracting 1 from both sides: \[x^2+4x=-1\] +The second step is the move the constant term to the right side of the equation by subtracting 1 from both sides: \[x^2+4x+1-1=0-1\] \[x^2+4x=-1\] -Our value for $b$ is 4. $4\div2 = 2$ and $2^2=4$, so we will add 4 to both sides of the equation (step three): \[x^2+4x=3\] +Our value for $b$ is 4. $4\div2 = 2$ and $2^2=4$, so we will add 4 to both sides of the equation (step three): \[x^2+4x+4=-1+4\] \[x^2+4x+4=3\] -The left side is now a perfect square. Because $x^2+4x=(x+2)^2$ we can rewrite it as a perfect square (step four): \[(x+2)^2=3\] +The left side is now a perfect square. Because $x^2+4x+4=(x+2)^2$ we can rewrite it as a perfect square (step four): \[(x+2)^2=3\] All that is left to do is to take the square root of both sides (step five): \[\sqrt{(x+2)^2}=\pm\sqrt{3}\] which gives us \[x+2=\pm\sqrt{3}\] -and then solve for $x$: \[x=\pm\sqrt{3} -2\] +and then solve for $x$ (step six): \[x=\pm\sqrt{3} -2\] Conventionally, we would write this as $x=-2+\sqrt{3}$ or $x=-2-\sqrt{3}$.