MHB What is the Relationship Between Roots and Coefficients in a Quadratic Equation?

[mathdad]

I found this question to be interesting.

View attachment 7495

It is probably never introduced in a precalculus course unless it is an honor precalculus course in high school or college.

 

[skeeter]

note ... $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha \cdot \beta = \dfrac{c}{a}$

$(\alpha + \beta)^2 = \dfrac{b^2}{a^2}$

$\alpha^2 + 2\alpha\beta + \beta^2 = \dfrac{b^2}{a^2}$

$\alpha^2 + \beta^2 = \dfrac{b^2}{a^2} - \dfrac{2c}{a} = \dfrac{b^2 - 2ac}{a^2}$

 

[HallsofIvy]

Skeeter showed that it is not something you must do but you certainly can do it that way.

With \alpha= \frac{-b+ \sqrt{b^2- 4ac}}{2a}, \alpha^2= \frac{b^2- 2b\sqrt{b^2- 4ac}+ b^2- 4ac}{4a^2}.

With \beta= \frac{-b- \sqrt{b^2- 4ac}}{2a}, \beta^2= \frac{b^2+ 2b\sqrt{b^2- 4ac}+ b^2- 4ac}{4a^2}.

Adding the two, the -2b\sqrt{b^2- 4ac} and 2b\sqrt{b^2- 4ac} cancel leaving

\alpha^2+ \beta^2= \frac{4b^2- 8ac}{4a^2}= \frac{b^2- 2ac}{a^2}