First slide

Theory of equations

Question

If $\tan θ_{1}, \tan θ_{2}, \tan θ_{3}$ are the real roots of the $x^{3} - (a + 1) x^{2} + (b - a) x - b = 0$ , where $θ_{1} + θ_{2} + θ_{3} \in (0, π)$ ,then $θ_{1} + θ_{2} + θ_{3}$ is equal to

Difficult

A

$π / 2$
B

$π / 4$
C

$3 π / 4$
D

$π$

Solution

$\begin{array}{l} \tan θ_{1} + \tan θ_{2} + \tan θ_{3} = (a + 1) \\ Σtan θ_{1} \tan θ_{2} = (b - a) \\ \tan θ_{1} \tan θ_{2} \tan θ_{3} = b \\ ∴ \tan (θ_{1} + θ_{2} + θ_{3}) = \frac{Σtan θ_{1} - Πtan θ_{1}}{1 - Σtan θ_{1} \tan θ_{2}} \\ = \frac{a + 1 - b}{1 - (b - a)} = 1 \\ \Rightarrow θ_{1} + θ_{2} + θ_{3} = \frac{π}{4} \end{array}$

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