Question

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

Answer 1

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 $b$.

\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)$$.