Question

Solve: \[{{x}^{2}}-\left( 3\sqrt{2}-2i \right)x-\sqrt{2}i=0\]

Answer 1

Hint:To solve the equation we use the complex number formula where we take the value of \[x\] as \[\left( a+bi \right)\] and then form the equation by separating the values and complex number values by solving the two equations up and down and finding the value of \[a\] and \[b\] and then placing the values in terms of \[x\].

Complete step by step solution:
According to the question given, the equation is \[{{x}^{2}}-\left( 3\sqrt{2}-2i \right)x-\sqrt{2}i=0\].
After writing the equation, we place the value of \[x\] in terms of \[\left( a+bi \right)\] and then we form the equation as:
\[\Rightarrow {{\left( a+bi \right)}^{2}}-\left( 3\sqrt{2}-2i \right)\left( a+bi \right)-\sqrt{2}i=0\]
Now we expand the equation and separate the values in terms of normal and complex number where we get the value of the equation as:
\[\Rightarrow \left( {{a}^{2}}-{{b}^{2}}-3\sqrt{2}a-2b \right)+i\left( -\sqrt{2}+2a+2ab-3\sqrt{2}b \right)\]
\[\Rightarrow 0+0i\]
Now writing the equation up and down so as to eliminate the values and then find the values in term of
\[a\] and \[b\].
\[\Rightarrow \begin{matrix}

\left( {{a}^{2}}-{{b}^{2}}-3\sqrt{2}a-2b \right)=0\text{ } \\
\left( -\sqrt{2}+2a+2ab-3\sqrt{2}b \right)=0 \\
\end{matrix}\]
Solving the two equations we find the value of \[a\] and \[b\] both positive and negative and then we find two equations of \[x\] with both being \[x=a\pm ib\]. Hence, after solving the equation we get the value of \[a=\pm \dfrac{3\sqrt{2}-4}{2}\] and \[b=\pm \dfrac{2+\sqrt{2}}{2}\].
Therefore, the two values of \[x\] is equal to \[\pm \dfrac{3\sqrt{2}-4}{2}\pm \dfrac{2+\sqrt{2}}{2}i\]

Note:
Complex number are number composed of real and imaginary number written in form of \[\left( a+bi
\right)\], students may go wrong if they try to solve it like a quadratic equation as the question will only get difficult to solve and get lengthy therefore, we first separate the equation in terms \[\left( a+bi \right)\].