# If The Sum Of The Roots Of The Quadratic Equation Ax2 + Bx + C = 0 Is Equal To The Sum Of The Squared Of Their Reciprocals Then A/C, B/A, C/B Are In:

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

The sum of roots are

$$\alpha + \beta = \frac{-b}{a} = \frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}\\\Rightarrow \frac{\alpha^{2} + \beta^{2}}{(\alpha \beta)^{2}} \\\Rightarrow \frac{(\alpha +\beta)^{2} – 2((\alpha \beta)}{(\alpha \beta)^{2}}$$

The product of the roots are αβ = $$\frac{c}{a}$$

Substituting α + β = $$\frac{-b}{a} and αβ = \frac{c}{a}$$, we get:

$$\frac{-b}{a} =\frac{(\frac{b}{a})^{2}-2(\frac{c}{a})}{(\frac{c}{a})^{2}}\\ \frac{-b}{a} = \frac{(\frac{b}{a})^{2}-2(\frac{c}{a})}{(\frac{c}{a})^{2}} = \frac{b^{2}-2ac}{c^{2}}$$

By Cross multiplying = $$bc^{2} = ab^{2} – 2ac$$

By Rearranging the terms = $$ab^{2} + bc^{2} = 2a^{2}c$$

By Dividing throughout by abc =$$\frac{b}{c} + \frac{c}{a} = \frac{2a}{b} \\\frac{b}{c},\frac{a}{b},\frac{c}{a}$$.

This is true if are in AP.

$$\frac{c}{b},\frac{b}{a},\frac{a}{c}$$. Hence their reciprocal, are in HP.

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