Let $\alpha$ and $\beta$ be the two roots of the quadratic equation \[ax^2+bx+c=0\;\;\;\;\;(a≠0)\] then, \[\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}\] \[\text{and,}\;\;\; \beta=\frac{-b-\sqrt{b^2-4ac}}{2a}\]
We have, \[\begin{array}{l} \alpha+\beta &= \frac{-b+\sqrt{b^2-4ac}-b-\sqrt{b^2-4ac}}{2a} \\ &= \frac{-2b}{2a} = \frac{-b}{a} \\ &= -\frac{\text{coefficient of }x}{\text{coefficient of }x^2} \end{array}\]
Again, \[\begin{array}{l} \alpha\cdot\beta &= \frac{(-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac})}{4a^2} \\ &= \frac{(-b)^2-(b^2-4ac)}{4a^2} \\ &= \frac{c}{a} \\ &= \frac{\text{constant term}}{\text{coefficient of }x^2} \end{array}\]
Thus, \[\begin{array}{l} \text{Sum of the roots} &= -\frac{b}{a} \\ \text{Product of the roots} &= \frac{c}{a} \end{array}\]
Special Roots
Under the following conditions, the given quadratic equation will have the special roots:
Case I: Roots equal in magnitude but opposite in sign
Two roots will be equal in magnitude but opposite in sign if their sum is zero. \[\begin{array}{l} \text{i.e.} & \alpha+\beta &= 0 \\ \text{i.e.} & -\frac{b}{a} &= 0 \\ \therefore & b &= 0 \end{array}\]
Case II: Reciprocal roots
The two roots will be reciprocal to each other if their product is $1$. \[\begin{array}{l} \text{i.e.} & \alpha\beta &= 1 \\ \text{i.e.} & \frac{c}{a} &= 1 \\ \therefore & c &= a\end{array}\]
Case III: One root zero
If one root is zero, then product of the roots is zero. \[\begin{array}{l} \text{i.e.} & \alpha\beta &= 0 \\ \text{i.e.} & \frac{c}{a} &= 0 \\ \therefore & c &= 0 \end{array}\]
Case IV: Both roots zero
If both roots are zero, then the sum and the product of the roots are zero. \[\begin{array}{l} \text{i.e.} & \alpha+\beta &= 0 \\ \text{i.e.} & -\frac{b}{a} &= 0 \\ \therefore & b &= 0 \end{array}\] \[\begin{array}{l} \text{and,} & \alpha\beta &= 1 \\ \text{i.e.} & \frac{c}{a} &= 1 \\ \therefore & c &= a\end{array}\]
Formation of a Quadratic Equation
Let $\alpha$ and $\beta$ be the two roots of the quadratic equation \[ax^2+bx+c=0\;\;\;\;\;(a≠0)\] then, \[\alpha+\beta = -\frac{b}{a}\] \[\text{and,}\;\;\; \alpha\beta=\frac{c}{a}\]
Now, \[\begin{array}{l} & ax^2+bx+c &= 0 \\ \text{or,} & x^2+\frac{b}{a}x+\frac{c}{a} &= 0 \\ \text{or,} & x^2-\left(-\frac{b}{a}\right)x+\frac{c}{a} &= 0 \\ \therefore & x^2-(\alpha+\beta)x+\alpha\beta &= 0\end{array}\]
Symmetric Functions of Roots
The functions: the sum $(\alpha+\beta)$ and the product $(\alpha\beta)$ are unaltered when the roots are interchanged. Hence, $\alpha+\beta$ and $\alpha\beta$ are the symmetric functions of $\alpha$ and $\beta$.
Thus, symmetric functions of the roots of a quadratic equation may be defined as those functions in which the two roots are so involved that the function is unaltered when the two roots are interchanged. Some examples of symmetric functions are \[\frac{1}{\alpha},\;\; \alpha^2+\beta^2,\;\; \frac{\alpha+\beta}{\alpha\beta},\;\; \alpha^2\beta^2,\;\; \text{etc.}\]
One of the interesting characteristics of symmetric functions in $\alpha$ and $\beta$ is that they can be expressed in terms of $\alpha+\beta$ and $\alpha\beta$. Hence, a symmetric function in $\alpha$ and $\beta$ can be expressed in terms of the coefficients of the equation.
Let $\alpha$ and $\beta$ be the roots of the quadratic equation \[x^2-px+q=0\] then, \[\alpha+\beta=p\] \[\text{and,} \alpha\beta=q\] From these values, it is easy to show that
\[\begin{array}{l} \text{1.} & \alpha^2+\beta^2 &= (\alpha+\beta)^2-2\alpha\beta \\ & & = p^2-2q \\ \text{and,} & \alpha^2\beta^2 &= q^2 \\ \text{2.} & \alpha^3+\beta^3 &= (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta) \\ & & = p(p^2-2q-q) \\ & & = p(p^2-3q) \\ \text{and,} & \alpha^3\beta^3 &= q^3 \end{array}\]
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