Quadratic Formula or Not?

There was a very interesting post in the AP Calc TEACHERS – AB/BC Facebook page last week. It discussed an apparently new method for solving quadratic equation without guess-and-check factoring, completing the square, or the quadratic formula. It was discovered by Dr. Po-Shen Loh of Carnegie Mellon University. The full article is here. The article links to a video of Dr. Loh explaining his method.

This method will work for any and all quadratic equations and is easy, almost intuitive, to understand. Folks have been solving quadratics in the West since long before Euclid’s time, in Muslin countries, in India, and in China. It is a surprise no one found this method before.

I’ll show you the method (an algorithm really). It is based on long known fact that for a quadratic equation in the form

{{x}^{2}}+px+q=0

the sum of the roots is –p, the opposite of the coefficient of the first degree-term, and the constant term, q, is the product of the root.

Example 1 (A simple example to show the method): Solve  {{x}^{2}}+6x-16=0

  • The sum of the roots is –6.
  • Therefore, the average of the roots is –3
  • Since there are only 2 roots, they must be the same distance of either side of –3. Let the roots be \left( {-3+u} \right) and \left( {-3-u} \right) with \displaystyle u\ge 0.
  • The product of the roots is \left( {-3+u} \right)\left( {-3-u} \right)=-16
  • Solving this equation:

9-{{u}^{2}}=-16

{{u}^{2}}=25

u=5

  • Then the roots are –3 + 5 = 2 and –3 – 5 = –8

Example 2 (More complicated, but no more difficult) Solve 3{{x}^{2}}-7x+8=0

  • The leading coefficient must be 1, so divide by 3: {{x}^{2}}-\frac{7}{3}x+\frac{8}{3}=0
  • The average of the roots is 7/6.
  • The roots are \left( {\frac{7}{6}-u} \right) and \left( {\frac{7}{6}+u} \right)
  • Solving

\displaystyle \left( {\frac{7}{6}+u} \right)\left( {\frac{7}{6}-u} \right)=\frac{8}{3}

\displaystyle \frac{{49}}{{36}}-{{u}^{2}}=\frac{8}{3}

\displaystyle {{u}^{2}}=\frac{{49}}{{36}}-\frac{8}{3}=-\frac{{47}}{{36}}

\displaystyle u=\frac{{\sqrt{{47}}}}{6}i

  • And the roots are  \displaystyle \frac{7}{6}+\frac{{\sqrt{{47}}}}{6}i  and  \displaystyle \frac{7}{6}-\frac{{\sqrt{{47}}}}{6}i

You may even prove the quadratic formula using this method. I’ll leave that for something to do over vacation – Don’t worry, it won’t take long.

Happy Holidays to everyone.


BTW My next posts will  be in January about Units 9 and 10 of the AP Calculus Course Description.

 

 

 

 


 

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