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# If x is real root, what is 'a' in the equation (2+i)x^2+(a-1)x+4-2i=0

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If x is real root, what is 'a' in the equation (2+i)x^2+(a-1)x+4-2i=0

math complex-numbers
jdh3010  Aug 10, 2015

#10
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Here's a more general approach (there are an infinite number of solutions, but only two real ones):

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Alan  Aug 10, 2015
#1
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I got

a=11/12 +ki

But i have not checked it and I would be extremely surprised if it is correct :)

Melody  Aug 10, 2015
#2
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Melody  Aug 10, 2015
#3
+94120
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Ok I think I have it now

(2+i)x^2+(a-1)x+4-2i=0

$$\\(2+i)x^2+(a-1)x+4-2i=0\\\\ 2x^2+(a-1)x+4+(x^2-2)i=0\\\\ 2x^2+(a-1)x+4=0 \qquad AND \qquad x^2-2=0\\\\ 2x^2+(a-1)x+4=0 \qquad AND \qquad x=\pm\sqrt2\\\\ NOW consider\\\\ 2x^2+(a-1)x+4=0\\\\ 2x^2+4=-(a-1)x\\\\$$

$$\\2*2+4=\pm\sqrt2(a-1)\\\\ 8=\pm\sqrt2(a-1)\\\\ \frac{\pm 8}{\sqrt2}=a-1\\\\ \pm 4\sqrt2=a-1\\\\ a=1\pm 4\sqrt2\\\\$$

$$\\If \;\;x=+\sqrt2\qquad then\qquad a=1-4\sqrt2\\\\ If\;\; x=-\sqrt2\qquad then\qquad a=1+4\sqrt2\\\\$$

Melody  Aug 10, 2015
#4
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You could also try a = (-97 + 49i)/5

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Alan  Aug 10, 2015
#5
+94120
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If x is real root, what is 'a' in the equation

(2+i)x^2+((-97 + 49i)/5-1)x+4-2i=0

$$\\(2+i)x^2+(\frac{(-97 + 49i)}{5}-1)x+4-2i=0\\\\ 2x^2+ix^2-\frac{97x}{5}+\frac{49xi}{5}-x+4-2i=0\\\\ 2x^2-x+4-\frac{97x}{5}+\frac{49xi}{5}-2i+ix^2=0\\\\ 10x^2-102x+20+\left(5x^2+49x-10\right)i=0\\\\$$

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{102}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{10}}\\ {\mathtt{x}} = {\frac{{\mathtt{1}}}{{\mathtt{5}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{10}}\\ {\mathtt{x}} = {\mathtt{0.2}}\\ \end{array} \right\}$$

$${\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{49}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{10}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{\mathtt{1}}}{{\mathtt{5}}}}\\ {\mathtt{x}} = -{\mathtt{10}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.2}}\\ {\mathtt{x}} = -{\mathtt{10}}\\ \end{array} \right\}$$

There does not seem to be a single common x value there Alan.

Am I thinking about it wrong, or did I make an error.  (I have not checked my working - I suppose I should.)

EDITED

Thanks Alan - Now it works :)

x=0.2       a = (-97 + 49i)/5

Melody  Aug 10, 2015
#6
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Your fourth line should have (5x2 + 49x - 10)i Melody, not (x2 + 49x - 10)i

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Alan  Aug 10, 2015
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Thanks Alan, now it works.

How did you get your answer and are you sure that there are no others?

Melody  Aug 10, 2015
#8
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This is how I found it (with the help of Mathcad):

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Alan  Aug 10, 2015
#9
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Thanks Alan,

I tried to do it this way with the help of Wolfram|Alpha but I didn't get an answer.

I might have been doing it wrong.

Also,

When you did it with Mathcad, it doesn't seem to have thrown up my answer.  Am I wrong?  Why was that?

Melody  Aug 10, 2015
#10
+27229
+10

Here's a more general approach (there are an infinite number of solutions, but only two real ones):

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Alan  Aug 10, 2015
#11
+94120
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Thanks Alan :)

Melody  Aug 10, 2015
#12
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The answer was actually 6 or -6

(2+i)x^2+(a-i)x+4-2i=0

→(2x^2+ax+4)+(x^2-x-2)i=0

x=2 or -1

substitute 2 in 2x^2+ax+4=0

8+2a+4=0, a=-6

substitue -1 in 2x^2+ax+4=0

2-a+4=0, a=6

So a is 6 or -6

jdh3010  Aug 10, 2015
#13
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That's the answer to a different question!

In your original question you had (a - 1)x.  In your latest post you have (a - i)x.

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Alan  Aug 10, 2015
#14
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Sorry, and thanks for correcting my error!!!

jdh3010  Aug 10, 2015
#15
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No problem.  We all make mistakes!  At least yours led to an interesting few analyses!

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Alan  Aug 10, 2015