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Question: Solve: \({x^2} - \left( {3\sqrt 2 + 2i} \right)x + 6i\sqrt 2 = 0\)...

Solve: x2−(32+2i)x+6i2=0{x^2} - \left( {3\sqrt 2 + 2i} \right)x + 6i\sqrt 2 = 0

Explanation

Solution

Hint: - Factorize the quadratic equation
Given equation is
x2−(32+2i)x+6i2=0 ⇒x2−32x−2ix+6i2=0 ⇒x(x−32)−2i(x−32)=0 ⇒(x−32)(x−2i)=0 ⇒(x−32)=0, (x−2i)=0 ⇒x=32 and x=2i  {x^2} - \left( {3\sqrt 2 + 2i} \right)x + 6i\sqrt 2 = 0 \\\ \Rightarrow {x^2} - 3\sqrt 2 x - 2ix + 6i\sqrt 2 = 0 \\\ \Rightarrow x\left( {x - 3\sqrt 2 } \right) - 2i\left( {x - 3\sqrt 2 } \right) = 0 \\\ \Rightarrow \left( {x - 3\sqrt 2 } \right)\left( {x - 2i} \right) = 0 \\\ \Rightarrow \left( {x - 3\sqrt 2 } \right) = 0,{\text{ }}\left( {x - 2i} \right) = 0 \\\ \Rightarrow x = 3\sqrt 2 {\text{ and }}x = 2i \\\
So, this is the required solution of the given equation.

Note: - In such types of questions the key concept is that we have to factorize the given equation, then we will get the required answer.