Question

If $\frac{1}{2-i}$ is a root of $a{x}^2+bx+c=0$ , where $i=\sqrt{-1}$ and $\frac{1}{3-2\sqrt{2}}$ is a root of $p{x}^2+dx+q=0$, then

• A. $a<b<d<c$
• B. $d<a<b<c$
• C. $d<b<c<a$
• D. $b<d<c<a$

Answer 1

Answer: (C)

$d<b<c<a$

$a{x}^2+bx+c=0$
One root is $\frac{1}{2-i}\times \frac{2+i}{2+i}=\frac{2+i}{5}$
Other root will be $\left(\frac{2-i}{5}\right)$
$x^2-\left(\frac{2+i+2-i}{5}\right)x+\left(\frac{2-i}{5}\right)\left(\frac{2+i}{5}\right)=0$
$5x^2-4x+1=0$
$a=5,b=-4,c=1$
$p{x}^2+dx+q=0$
One root is $\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}=(3+2\sqrt{2})$
Then, the other root will be $3-2\sqrt{2}$
$x^2-(3+2\sqrt{2}+3-2\sqrt{2})x+(3+2\sqrt{2})(3-2\sqrt{2})=0$
$x^2-6x+1=0$
$p=1,d=-6,q=1$
Hence, $D$ is correct.