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

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MarkFL

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Feb 24, 2012
13,735
As a student, I was taught 3 ways to solve quadratic equations:

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:

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MarkFL

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Feb 24, 2012
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Here is another method submitted to me by our own agentmulder:

Let

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

\(\displaystyle ax^2+bx=-c \)

Now... I want to complete the square but the coefficient of x^2 is bothering me. NO PROBLEM, I'll take it's square root.

\(\displaystyle \left(\sqrt{a}x+? \right)^2 \)

Now... what is the question mark? A little playing around shows it must be \(\displaystyle \frac{b}{2 \sqrt{a}}\) because that's the only way to get $bx$ when we square the binomial, and I'll just subtract (it's square), the constant as a correction term.

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

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

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

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

DIVIDE BY $\sqrt{a}$ AND YOU'RE DONE!

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