#1**+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

#3**+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

#5**+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**+5 **

Your fourth line should have (**5**x^{2} + 49x - 10)i Melody, not (x^{2} + 49x - 10)i

.

Alan
Aug 10, 2015

#7**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

#9**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**+10 **

Best Answer

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

.

Alan
Aug 10, 2015

#12**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