Question

# Let $$x_1, x_2,$$ are the roots of quadratic equation $$x^2 + ax + b = 0$$, Where $$a, b$$ are complex numbers and $$y_1, y_2$$ are the roots of the quadratic equation $$y^2 + \left|a\right| y + \left|b\right| =0$$. If $$\left|x_1\right| = \left|x_2\right| = 1$$, then

A
|y1|=|y2|=1.
B
|y1|=|y2|1.
C
|y1|1,|y2|=1.
D
|y1|=1,|y2|1.

Solution

## The correct option is A $$\left|y_1\right| = \left|y_2\right|= 1$$.Now $$x_{1}.x_{2}=b$$Hence$$|x_{1}.x_{2}|$$$$=|x_{1}|.|x_{2}|$$$$=1.1$$$$=1$$$$=|b|$$ ...(i)And $$-a=x_{1}+x_{2}$$$$|a|=|x_{1}+x_{2}|$$Now applying triangle inequality, we get $$|x_{1}+x_{2}|\leq |x_{1}|+|x_{2}|$$Hence$$|a|\leq |x_{1}|+|x_{2}|$$Hence$$|a|\leq 2$$Considering $$|a|=2$$ we get $$y^{2}+2y+1=0$$$$(y+1)^{2}=0$$$$y=-1,-1$$Hence$$|y_{1}|=|y_{2}|=1$$.Mathematics

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