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

\[ax^2+bx+c=0\Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0\]
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.
I encountered this method when I was trying to solve quadratic congruences of prime moduli in number theory (see >>this<<).
Kind Regards,
Sudharaka.
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 by a:
$\displaystyle x^2+\frac{b}{a}x=-\frac{c}{a}$
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 for x:
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
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 for x:
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
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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