Question

Find the value of
i) x²+2x2-3x+21, if x = 1+2i.​

Answer 1

Step-by-step explanation:

Given, x=1+2i

therefore,

x^2+2x2-3x+21

=(1+2i)^2+2(1+2i)2-3(1+2i)+21

=1+4i+(2i)^2+4+8i-3-6i+21

=1+4i-4+4+8i-3-6i+21 [i^2=-1]

=19+6i

That's may be your answer but I am

not sure because I can't understand what do you mean by 2x2 .Here I consider it as the letter for x=1+2i. So I'm not sure about it. I hope it will help you

Answer 2

The required value is 9+6i

Step-by-step explanation:

The given expression is:

$x^{2} +2x^{2} -3x+21$

Where $x=1+2i$

First simplify the given expression

$=3x^{2} -3x+21\\\\=3(x^{2} -x)+21$

Now substitute the value of x

$=3((1+2i)^{2}-(1+2i))+21 \\\\=3(1^{2} +(2i)^{2} +2\times 1\times 2i-1-2i)+21$

we know that

$i=\sqrt{-1} \\\\\Rightarrow i^{2}=-1$

So , by plugging it we get.

$=3(1-4+4i-1-2i)+21\\=3(-4+2i)+21\\=-12+6i+21\\=9+6i$

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