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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

Best Answer 

 #10
avatar+26322 
+10

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

 solutions for a

.

Alan  Aug 10, 2015
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15+0 Answers

 #1
avatar+90988 
+5

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
avatar+90988 
+5

Thanks for your generosity jdh, but my answer is nonsense.  

Melody  Aug 10, 2015
 #3
avatar+90988 
+10

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
avatar+26322 
+5

You could also try a = (-97 + 49i)/5

.

Alan  Aug 10, 2015
 #5
avatar+90988 
+5

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
avatar+26322 
+5

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

.

Alan  Aug 10, 2015
 #7
avatar+90988 
0

Thanks Alan, now it works.

I have edited my check of your answer.

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

Melody  Aug 10, 2015
 #8
avatar+26322 
+5

This is how I found it (with the help of Mathcad):

 the value of a

 

.

Alan  Aug 10, 2015
 #9
avatar+90988 
0

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
avatar+26322 
+10
Best Answer

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

 solutions for a

.

Alan  Aug 10, 2015
 #11
avatar+90988 
0

Thanks Alan :)

Melody  Aug 10, 2015
 #12
avatar+299 
0

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
avatar+26322 
+5

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.

 

.

Alan  Aug 10, 2015
 #14
avatar+299 
+5

I've made a mistake.

Sorry, and thanks for correcting my error!!!

jdh3010  Aug 10, 2015
 #15
avatar+26322 
0

No problem.  We all make mistakes!  At least yours led to an interesting few analyses!

.

Alan  Aug 10, 2015

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