Question

If $\frac{1}{2-i}$ is a root of $a{x}^2+bx+c=0$ 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

The correct option is C $d<b<c<a$

$a{x}^2+bx+x=0$--------------------(1)

has a root $\frac{1}{2-i}=\frac{2+i}{(2-i)(2+i)}=\frac{2+i}{5}$

such complex roots comes in pair, hence the other root will be the conjugate of the previous i.e. $\frac{2-i}{5}$

$\therefore$ The equation will be

$(x-\frac{2+i}{5})(x-\frac{2-i}{5})=0$

$\Rightarrow x^2-x\left(\frac{4}{5}\right)+\left(\frac{1}{5}\right)=0$

$\Rightarrow 5x^2-4x+1=0$-------------------(2)

Comparing Equation(1) & (2), we get

$a=5,b=-4,c=1$

Also,

$p{x}^2+dx+q=0$------------(3)

has a root $\frac{1}{3-2\sqrt{2}}=\frac{3+2\sqrt{2}}{(3-2\sqrt{2})(3+2\sqrt{2})}=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$

The other root will be $3-2\sqrt{2}$

$\therefore$ The Equation with roots $3+2\sqrt{2}$ and $3-2\sqrt{2}$ will be

$x^2-\left(6\right)x+1=0$----------(4)

Comparing Equation (3) & (4), we get

$p=1$, $d=-6$, $q=1$

$\therefore d<b<c<a$