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Let ^@\alpha^@ and ^@\beta^@ be the roots of ^@x^2 - 3 x + c = 0^@, where ^@c^@ is a real number. If ^@-\alpha^@ is a root of ^@x^2 + 3 x - c = 0^@, find the value of ^@\alpha\beta^@.
Answer:
^@0^@
- ^@\alpha \space \& \space \beta ^@ are the roots of the equation, therefore,
^@\alpha \beta = \dfrac{c}{1} = c \space\space .....(1)^@ - As ^@\alpha^@ is the root of the equation ^@x^2 - 3 x + c = 0^@,
^@\alpha^2 - 3\alpha + c = 0 \space \space .....(2)^@
Also, ^@-\alpha^@ is the root of the equation ^@x^2 + 3 x - c = 0^@,
^@\begin{align} & (-\alpha)^2 + 3(-\alpha) - c = 0\\ &\implies \alpha^2 - 3\alpha - c = 0 && .....(3) \end{align}^@ - On subtracting ^@eq(2)^@ by ^@eq(3)^@, we get,
^@\begin{align} & \alpha^2 - 3\alpha - c - (\alpha^2 - 3\alpha +c) = 0 \\ \implies & \alpha^2 - 3\alpha - c - \alpha^2 + 3\alpha - c = 0 \\ \implies & -2c = 0 \\ \implies & c = 0 \end{align}^@ - By ^@eq(1)^@, we have,
^@\begin{align} & \alpha\beta = c \\ \implies & \alpha\beta = 0 \end{align}^@
