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So I solved the equation: $|x| + 3x = 2 + 6i$ but I don't know where my error is. And I know there is an error because Wolfram Alpha shows that the only solution is $ x = 2i $ . In my calculations I have 2 solutions.

$$|x| + 3x = 2 + 6i$$ $$\sqrt{a^2 + b^2} +3(a + bi) = 2 +6i$$ $$\sqrt{a^2 + b^2} = 2 +6i - 3a -3bi$$ $$\sqrt{a^2 + b^2} = 2-3a +(6 -3b)i$$ $$6 -3b=0$$ $$b=2$$ $$\sqrt{a^2 + 4} = 2-3a $$ $$a^2 + 4 = 4-12a+9a^2 $$ $$8a^2-12a=0 $$ $$4a(2a-3)=0 $$ $$a = 0 || a=3/2 $$

Why $x = 3/2 + 2i$ is not an answer?

Cotton Headed Ninnymuggins
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cocacola
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1 Answers1

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When you square both sides of $\sqrt{a^2+4} = 2-3a$, an extra solution is generated.

In particular, the solution $a = 3/2$ is the solution to $\sqrt {a^2+4} = 3a-2$, which is a perfectly valid solution for the squared equation $a^2+4 = (2-3a)^2$.

However for this $a$, $\sqrt{a^2+4} > 0$ but $2-3a < 0$, rendering it invalid for the original equation. For this reason, we must check whether all our solutions satisfy the original equation. There are no invalid steps in your calculations otherwise.

player3236
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  • Oh, thank you Sir! That's right. It is because square function gives same solution for 2 different arguments. In those cases do I need to check my equation for every solution or is there any other way to eliminate wrong answers? – cocacola Nov 01 '20 at 17:55
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    I don't think there is a way to eliminate unwanted solutions beforehand. – player3236 Nov 01 '20 at 17:59