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As a student, I was taught 3 ways to solve quadratic equations:

(1) $\displaystyle ax^2+bx+c=0$

To complete the square, I was taught to move the constant term to the other side and divide through by

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

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

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

And we have the famous quadratic formula.

In my years on math forums, I have gleaned two variations on this technique that I would like to share:

Method 1:

Divide (1) by

$\displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0$

Now, we next want to shift the roots to the right by 1/2 the value of the coefficient of the linear term, so our new equation is:

$\displaystyle \left(x-\frac{b}{2a} \right)^2+\frac{b}{a}\left(x-\frac{b}{2a} \right)+\frac{c}{a}=0$

$\displaystyle x^2-\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{b}{a}x-\frac{b^2}{2a^2}+\frac{c}{a}=0$

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

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

Now, we subtract $\displaystyle \frac{b}{2a}$ from these roots, to get the roots of the original equation:

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

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

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

Comments and questions should be posted here:

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

*i*) Factoring*ii*) Completing the square*iii*) Applying the quadratic formula, derived by completing the square on the general quadratic in standard form:(1) $\displaystyle ax^2+bx+c=0$

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.

In my years on math forums, I have gleaned two variations on this technique that I would like to share:

Method 1:

Divide (1) by

*a*:$\displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0$

Now, we next want to shift the roots to the right by 1/2 the value of the coefficient of the linear term, so our new equation is:

$\displaystyle \left(x-\frac{b}{2a} \right)^2+\frac{b}{a}\left(x-\frac{b}{2a} \right)+\frac{c}{a}=0$

$\displaystyle x^2-\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{b}{a}x-\frac{b^2}{2a^2}+\frac{c}{a}=0$

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

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

Now, we subtract $\displaystyle \frac{b}{2a}$ from these roots, to get the roots of the original equation:

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

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

Comments and questions should be posted here:

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

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