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Show work and/or describe how the expression for the completing the square method and the expression associated with the quadratic formula are equivalent.

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Given a general quadratic expression:


ax^2+bx+c=0

firs, lets divide both sides of the equation by 'a' :


\begin{gathered} ((1)/(a))(ax^2+bx+c=0)^{} \\ \Rightarrow(a)/(a)x^2+(b)/(a)x+(c)/(a)=0 \\ \Rightarrow x^2+(b)/(a)x+(c)/(a)=0 \end{gathered}

next, we can move the term c/a to the right side of the equation:


\begin{gathered} x^2+(b)/(a)x+(c)/(a)=0 \\ \Rightarrow x^2+(b)/(a)x=-(c)/(a) \end{gathered}

now we are ready to complete the square on the left side. What we have to do, is to take the constant that is multiplying x (in this case,b/a), and first, we divide it by 2, and then elevate to the square the result:


\begin{gathered} (b)/(a)(\cdot)/(\cdot)2=(b)/(2a) \\ \Rightarrow((b)/(2a))^2=(b^2)/(4a^2) \end{gathered}

then, adding this number on both sides of the equation, we get:


x^2+(b)/(a)x+(b^2)/(4a)=-(c)/(a)+(b^2)/(4a^2)

which we can write like this:


(x+(b)/(2a))^2=(-4ac+b^2)/(4a^2)_{}

applying the square root on both sides,we get the following:


\begin{gathered} \sqrt[]{(x+(b)/(2a))^2}=\sqrt[]{(b^2-4ac)/(4a^2)}=\pm\frac{\sqrt[]{b^2_{}-4ac}}{2a} \\ \Rightarrow x+(b)/(2a)=\pm\frac{\sqrt[]{b^2-4ac}}{2a} \end{gathered}

finally, we can solve for x:


\begin{gathered} x+(b)/(2a)=\pm\frac{\sqrt[]{b^2-4ac}}{2a} \\ \Rightarrow x=-(b)/(2a)\pm\frac{\sqrt[]{b^2-4ac}}{2a}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}

as we can see, if we have a general quadratic equation, we can us the completing the square method to deduce the quadratic formula

User Edthrn
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