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Kenneth John Odle 11 months ago
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1. 17
complete-the-square.tex

#### 17 complete-the-square.tex View File

 @ -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 forx$in equation \ref{eq2} will result in one of   For equations that are not easily factored, a general solution called the quadratic equation'' can be used to solve any quadratic.   For any equation of the form$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}\$.