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**Commentary for "Two methods for deriving the quadratic formula that I was not taught in school"**

Hi MarkFL,Moderator edit: This topic is for commentary pertaining to the tutorial:

Two methods for deriving the quadratic formula that I was not taught in school

We were taught(or maybe something that I picked up somewhere) to think about the completion of the square in the following manner.To complete the square, I was taught to move the constant term to the other side and divide through bya:

$\displaystyle x^2+\frac{b}{a}x=-\frac{c}{a}$

Then, add the square of one-half the coefficient of the linear term to both sides:

$\displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a} \right)^2=-\frac{c}{a}+\left(\frac{b}{2a} \right)^2$

Write the left side as a square, and combine terms on the right:

$\displaystyle \left(x+\frac{b}{2a} \right)^2=\frac{b^2-4ac}{(2a)^2}$

Apply the square root property:

$\displaystyle x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$

Solve forx:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

And we have the famous quadratic formula.

\[ax^2+bx+c=0\Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0\]

Now the first two terms could be made up by squaring \(x+\frac{b}{2a}\). But when we square it we get an additional term, \(\frac{b^2}{4a^2}\). Therefore,

\[x^2+ \frac{b}{a}x+ \frac{c}{a}=\left(x+\frac{b}{2a} \right)^2- \frac{b^2}{4a^2}+ \frac{c}{a}=0\]

Simplifying this we get the quadratic formula.

Method 2:

Arrange (1) as:

$\displaystyle ax^2+bx=-c$

Multiply by $\displaystyle 4a$:

$\displaystyle 4a^2x^2+4abx=-4ac$

Add $\displaystyle b^2$ to both sides:

$\displaystyle 4a^2x^2+4abx+b^2=b^2-4ac$

Write the left side as a square:

$\displaystyle (2ax+b)^2=b^2-4ac$

Apply the square root property:

$\displaystyle 2ax+b=\pm\sqrt{b^2-4ac}$

Solve forx:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

I encountered this method when I was trying to solve quadratic congruences of prime moduli in number theory (see >>this<<).

Kind Regards,

Sudharaka.

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