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Relation Between Roots and Coefficient of Quadratic Equation - (Concept)

Let alpha and beta be two roots of a quadratic equation. So, we have

$\begin{aligned} & \alpha=\frac{-b-\sqrt{D}}{2 a} \\ & \beta=\frac{-b+\sqrt{D}}{2 a}\end{aligned}$

The sum of roots:

$\alpha+\beta=\frac{-\mathrm{b}-\sqrt{\mathrm{D}}}{2\mathrm{a}}+\frac{\mathrm{b}+\sqrt{\mathrm{D}}}{2 \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{a}}$

Product of roots:

$\begin{aligned} & \alpha \cdot \beta=\left(\frac{-\mathrm{b}-\sqrt{\mathrm{D}}}{2 \mathrm{a}}\right) \cdot\left(\frac{-\mathrm{b}+\sqrt{\mathrm{D}}}{2\mathrm{a}}\right)\\&=\frac{\mathrm{b}^2-\mathrm{D}}{4\mathrm{a}^2}=\frac{\mathrm{b}^2-\mathrm{b}^2+4\mathrm{ac}}{4\mathrm{a}^2}=\frac{4 \mathrm{ac}}{4 \mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}}\end{aligned}$

The difference of root can also be found in the same way by manipulating the terms

$\alpha-\beta=\left|\frac{\sqrt{D}}{a}\right|$

Important Results

(i) $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$
(ii) $\alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta)$
(iii) $\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$
(iv) $\alpha^3-\beta^3=(\alpha-\beta)^3+3 \alpha \beta(\alpha-\beta)$

Exam Chapter
JEE MAIN Complex numbers and quadratic equations
Algebra (Arihant)
Page No. : 108
Line : 1

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