Question

If $3+4i$ is a root of $x^2+Ax+B=0$ and $\sqrt{3}-2$ is a root of $x^2+Cx+D=0$, where $A,B,C$ and $D$ are all rational numbers, then

• A. $A<C<D<B$
• B. $A<D<C<B$
• C. $A>C>D>B$
• D. $A>D>C>B$

Answer 1

Answer: (B)

$A<D<C<B$

Consider $x^2+Ax+B=0$
It is given that both A and B are rational numbers and one root is $3+4i$.
Since the coefficients are rational, hence the complex roots will occur in conjugate pair.
Hence the other root will be $3-4i$.
Thus $A=-\left[\right(3+4i)+(3-4i\left)\right]$
$A=-6$ ...(i)
And
$B=(3+4i)(3-4i)$
$=25$ ...(ii).
Similarly consider the second equation.
$x^2+Cx+D=0$
It is given that one root is $\sqrt{3}-2$. Since both C and D are rational and non zero, hence the other root will be $-2-\sqrt{3}$
Therefore
$C=-\left[\right(\sqrt{3}-2)+(-\sqrt{3}-2\left)\right]$
$=4$ ...(iii)
And
$D=(\sqrt{3}-2)(-\sqrt{3}-2)$
$=-(\sqrt{3}-2)(\sqrt{3}+2)$
$=-(3-4)$
$=1$...(iv)
Hence
$A<D<C<B$ from i ii iii and iv.