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

AI Thread Summary
The discussion presents two alternative methods for deriving the quadratic formula beyond the traditional techniques of factoring, completing the square, and direct application. The first method involves manipulating the standard form of the quadratic equation to shift the roots and express it as a square, ultimately leading to the quadratic formula. The second method simplifies the equation by multiplying through by 4a, completing the square, and then solving for x. Additionally, a third method is introduced that utilizes the square root of the leading coefficient to facilitate the completion of the square. These variations provide valuable insights for students seeking deeper understanding of quadratic equations.
MarkFL
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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:

http://www.mathhelpboards.com/f49/commentary-two-methods-deriving-quadratic-formula-i-not-taught-school-4225/#post11756
 
Last edited:
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Here is another method submitted to me by our own agentmulder:

Let

$$ax^2+bx+c=0$$

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

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.

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

Now... what is the question mark? A little playing around shows it must be $$\frac{b}{2 \sqrt{a}}$$ 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.

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

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

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

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

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

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