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. $\left|y_1\right|=\left|y_2\right|=1$.
• B. $\left|y_1\right|=\left|y_2\right|\ne 1$.
• C. $\left|y_1\right|\ne 1,\left|y_2\right|=1$.
• D. $\left|y_1\right|=1,\left|y_2\right|\ne 1$.

Answer 1

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
$|x_1+x_2|\le |x_1|+|x_2|$
Hence
$|a|\le |x_1|+|x_2|$
Hence
$|a|\le 2$
Considering
$|a|=2$ we get
$y^2+2y+1=0$
$(y+1{)}^2=0$
$y=-1,-1$
Hence
$|y_1|=|y_2|=1$.