Quadratic EquationHard
Question
If the roots of quadratic equation $ax^{2} + bx + c = 0$ are $\frac{m}{m\ 1}$ and $\frac{m + 1}{m}$, then $(a + b + c)^{2}$ is equal to
Options
A.$b^{2} - ac$
B.$4\left( b^{2} - 4ac \right)$
C.$b^{2} - 4ac$
D.$\frac{b^{2} + c^{2}}{a^{2}}$
Solution
$ax^{2} + bx + c = a(x - \alpha)(x - \beta)$
$$\therefore\ \begin{matrix} a + b + c & \ = a(1 - \alpha)(1 - \beta) = a\left( 1 - \frac{m}{m - 1} \right)\left( 1 - \frac{m + 1}{m} \right) = \frac{a}{m(m - 1)} \\ |\alpha - \beta| & \ = \left| \frac{m}{m - 1} - \frac{m + 1}{m} \right| = \frac{1}{|m(m - 1)|} \\ (a + b + c)^{2} & \ = \frac{a^{2}}{(m(m - 1))^{2}} = a^{2}(\alpha - \beta)^{2} = a^{2}\left( (\alpha + \beta)^{2} - 4\alpha\beta \right) \\ & \ = a^{2}\left( \frac{b^{2}}{a^{2}} - \frac{4c}{a} \right) = b^{2} - 4ac \end{matrix}$$
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