Question

Solve the following quadratic equation:
\[{x^2} - (3\sqrt 2 - 2i)x - \sqrt 2 i = 0\]

Answer 1

Hint: we can use the Dharacharya formula in this question to find the roots of this quadratic equation. This question also has the involvement of iota. So one more thing to remember is that the value of ${i^2}$ is $ - 1$ . The formula of the discriminant which is very important is also used here which is required in the dharacharya formula.
Formula used: $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
$D = {b^2} - 4ac$

Complete step by step answer:
In the given figure,
\[{x^2} - (3\sqrt 2 - 2i)x - \sqrt 2 i = 0\]
Where $i$ stands for iota.
We know,
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where, $D = {b^2} - 4ac$ for a general quadratic equation $(a{x^2} + bx + c = 0)$
Now,
Let’s calculate D.
$D = {b^2} - 4ac$
On comparing, from given equation
$b = - (3\sqrt 2 - 2i)$
$a = 1$
$c = - \sqrt 2 i$
Put these values in the above equation,
$D = {\left( { - \left( {3\sqrt 2 - 2i} \right)} \right)^2} - 4\left( 1 \right)\left( { - \sqrt 2 i} \right)$
The value of ${\left( { - 1} \right)^2}$ is $1$
\[D = {\left( {3\sqrt 2 - 2i} \right)^2} + 4\sqrt 2i \]
Using identity ${\left( {a - b} \right)^2} = \left( {{a^2} + {b^2} - 2ab} \right)$
$D = {\left( {3\sqrt 2 } \right)^2} + {\left( {2i} \right)^2} - 2\left( {3\sqrt 2 } \right)\left( {2i} \right) + 4\sqrt 2 i$
$D = 18 + 4{i^2} - 12\sqrt 2 i + 4\sqrt 2 i$
The value of ${i^2}$ is $ - 1$.
So,
$D = 18 - 4 - 8\sqrt 2 i$
$D = 14 - 8\sqrt 2 i$
We can also write $14 = 16 - 2$
$D = 16 - 2 - 8\sqrt 2 i$
We can also write $16$ as ${\left( 4 \right)^2}$ and $ - 2$ as ${\left( {\sqrt 2 i} \right)^2}$
 $D = {\left( 4 \right)^2} + {\left( {\sqrt 2 i} \right)^2} - 8\sqrt 2 i$
Now,
Using the formula $\left( {{a^2} + {b^2} - 2ab} \right) = {\left( {a - b} \right)^2}$
$D = {\left( {4 - \sqrt 2 i} \right)^2}$
Now, using dharacharya formula
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
$x = \dfrac{{ - \left( { - \left( {3\sqrt 2 - 2i} \right)} \right) \pm \sqrt {{{\left( {4 - \sqrt 2 i} \right)}^2}} }}{{2 \times 1}}$
$x = \dfrac{{\left( {3\sqrt 2 - 2i} \right) \pm \left( {4 - \sqrt 2 i} \right)}}{2}$
$x = \dfrac{{\left( {3\sqrt 2 - 2i} \right)}}{2} \pm \dfrac{{\left( {4 - \sqrt 2 i} \right)}}{2}$
Therefore, these are the two values of x.

Note:
 complex numbers and quadratic equations are very much interrelated topics. So, while doing a question of one topic one can face difficulty if he does not know the concept of another topic. One more important thing to note here is that if we try to convert the value of D as a whole square of any number there is some variable or iota coming in it.