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

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The relationship between the roots and coefficients of a quadratic equation is defined by the formulas α + β = -b/a and α · β = c/a. These relationships are often not covered in standard precalculus courses but are crucial for understanding quadratic equations. The discussion highlights how to derive the sum of the squares of the roots, α^2 + β^2, using the coefficients. It shows that α^2 + β^2 can be expressed as (b^2 - 2ac)/a^2. This mathematical exploration emphasizes the interconnectedness of roots and coefficients in quadratic equations.
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I found this question to be interesting.

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

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

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

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

$\alpha^2 + \beta^2 = \dfrac{b^2}{a^2} - \dfrac{2c}{a} = \dfrac{b^2 - 2ac}{a^2}$
 
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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}
 
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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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