Question

The real values of x and y for which the equation $\left( {{x}^{4}}+2xi \right)-\left( 3{{x}^{2}}+yi \right)=\left( 3-5i \right)+\left( 1+2yi \right)$ is satisfied, are:
a). $x=2,y=3$
b). $x=-2,y=\dfrac{1}{3}$
c). Both (a) and (b)
d). None of these

Answer 1

Hint: First of all, separate and group all the real numbers in one and the complex numbers in the other on both sides of the equation. Then comparing on both sides of the equation, we will get conditions in x and y .Solving these conditions will give us the final answer.

Complete step-by-step solution -
Given equation is $\left( {{x}^{4}}+2xi \right)-\left( 3{{x}^{2}}+yi \right)=\left( 3-5i \right)+\left( 1+2yi \right)$
Removing all the brackets and simplifying,
$\begin{aligned} & {{x}^{4}}+2xi-3{{x}^{2}} - yi=3-5i+1+2yi \\\ & \Rightarrow \left( {{x}^{4}}-3{{x}^{2}} \right)+\left( 2x-y \right)i=4+\left( 2y-5 \right)i \\\ \end{aligned}$
Comparing on the both sides, we get,
${{x}^{4}}-3{{x}^{2}}=4$ and $2x-y=2y-5$
Let us consider the equation ${{x}^{4}}-3{{x}^{2}}=4$ .
Rearranging,
$\Rightarrow {{x}^{4}}-3{{x}^{2}}-4=0$
Now this can be written as
$\Rightarrow {{\left( {{x}^{2}} \right)}^{2}}-3{{x}^{2}}-4=0$ .
Thus this equation is a quadratic equation in ${{x}^{2}}$ . Therefore, its factors can be written as,
$\begin{aligned} & \Rightarrow {{\left( {{x}^{2}} \right)}^{2}}-4{{x}^{2}}+{{x}^{2}}-4=0 \\\ & \Rightarrow {{x}^{2}}\left( {{x}^{2}}-4 \right)+1\left( {{x}^{2}}-4 \right)=0 \\\ & \Rightarrow \left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-4 \right)=0 \\\ \end{aligned}$
Thus, we get, ${{x}^{2}}+1=0$ and ${{x}^{2}}-4=0$ .
Thus, the four roots are 2,-2 and i,–i. But since in the question it is told that x should be real, the permitted values are +2 and -2.
Now, let us consider the equation $2x-y=2y-5$ .
Rearranging,
$\begin{aligned} & \Rightarrow 2x+5=3y \\\ & \Rightarrow y=\dfrac{2x+5}{3} \\\ \end{aligned}$
Let us put x=2 at first place.
We get,
$\begin{aligned} & \Rightarrow y=\dfrac{2\left( 2 \right)+5}{3} \\\ & \Rightarrow y=\dfrac{4+5}{3} \\\ & \Rightarrow y=\dfrac{9}{3} \\\ & \Rightarrow y=3 \\\ \end{aligned}$
Thus one pair is x=2 and y=3.
Now, let us put x=-2.
$\begin{aligned} & \Rightarrow y=\dfrac{2\left( -2 \right)+5}{3} \\\ & \Rightarrow y=\dfrac{-4+5}{3} \\\ \end{aligned}$
$\Rightarrow y=\dfrac{1}{3}$.
Thus, we get one more pair, x=-2 and $y=\dfrac{1}{3}$.
Therefore, both options, option (a) and (b) are correct.
But option (c) says “Both (a) and (b)”. Thus, Option (c) is the correct option.

Note: Please be careful while solving the quadratic equation. It should give you four roots, maybe real or complex. If you get less than four roots then you have solved it wrongly. Some students assign ${{x}^{2}}$ to some variable (say z) and then forget to reassign z back to ${{x}^{2}}$ and get only roots. This would give you a completely wrong answer.