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Two equations $a_1z^2 + b_1z +c_1 = 0$ and $a_2z^2 + b_2z +c_2 = 0$ where $a_i,b_i,c_i \in \mathbb C , i\in \{1,2\}$ have exactly one root common which is a positive real number . Also it is know that the equation $a_1z^2 + (b_1-1)z +c_1 = 0$ and $a_2z^2 + (b_2-1)z +c_2 = 0$ also have exactly one common root ,which is another positive real ,then answer the following : $$1.\arg\left(\frac{b_1-b_2}{a_1-a_2}\right)$$ $$2.\arg\left(\frac{a_1b_2-a_2b_1}{a_1-a_2}+\frac{b_1c_2-b_2c_1}{c_2-c_1}\right)$$

My Work $$\frac{{{\alpha}_1}^2}{b_1c_2-b_2c_1}=\frac{-{\alpha}_1}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1} $$ $$\implies \frac{b_1c_2-b_2c_1}{a_1c_2-a_2c_1}=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}$$ Similarly $$\implies \frac{(b_1-1)c_2-(b_2-1)c_1}{a_1c_2-a_2c_1}=\frac{a_1c_2-a_2c_1}{a_1(b_2-1)-a_2(b_1-1)}$$ Dividing both the expressions $$\frac{b_1c_2-b_2c_1}{(b_1-1)c_2-(b_2-1)c_1}=\frac{a_1(b_2-1)-a_2(b_1-1)}{a_1b_2-a_2b_1}$$ $$\frac{c_2-c_1}{(b_1-1)c_2-(b_2-1)c_1}=\frac{a_2-a_1}{a_1b_2-a_2b_1}$$ $$\frac{a_1b_2-a_2b_1}{a_2-a_1}=\frac{b_1c_2-b_2c_1}{c_2-c_1}-1$$ $$\frac{a_1b_2-a_2b_1}{a_1-a_2}+\frac{b_1c_2-b_2c_1}{c_2-c_1}=1$$ But for first part I am not able to solve .

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  • $\begingroup$ Does #1 correspond to first equation pair while #2 the second? $\endgroup$ – Ng Chung Tak Nov 22 '16 at 17:30
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Hint: let $x_1 \ne x_2 \in \mathbb{R}^+$ be the common real positive roots of the two pairs of equations. Then:

$$ (a_1-a_2)x_1^2 + (b_1-b_2)x_1 + (c_1-c_2) = 0 $$ $$ (a_1-a_2)x_2^2 + (b_1-b_2)x_2 + (c_1-c_2) = 0 $$

Since $x_1 \ne x_2$ it follows that $x_1,x_2$ are the two distinct roots of:

$$ (a_1-a_2)z^2 + (b_1-b_2)z + (c_1-c_2) = 0 $$

which implies $a_1 \ne a_2$ and:

$$ x_1+x_2 = -\,\frac{b_1-b_2}{a_1-a_2} \;\;\in\;\;\mathbb{R}^+ $$

so $\arg\big(\frac{b_1-b_2}{a_1-a_2} \big) = \cdots$

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