Question

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

Answer 1

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