Melody

Hi Dbg,  

Jdh3010's graph is NOT a function because it does not pass the vertical line test.

When you draw any vertical line through a function it can cross the fuction in AT MOST one place.  

This graph would often have two crossing points :)

This means that some x values have more than one y value.      

If a graph or an equation is a function it means that no x value can have more than one y value.  

2*5-1-0=9  that is one answer.

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

Thanks Maia,

Here is a graph of y=tan(theta)

It illustrates what Geno is telling you  :)

cos(25)/21        should work fine

$${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}}{{\mathtt{21}}}} = {\mathtt{0.043\: \!157\: \!513\: \!668\: \!428\: \!6}}$$

Ok I think I have it now   

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

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

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

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

1,2,4,8,....2^(n-1).......2^29

a=1, r=2   n=30

$$\\S_n=\frac{a(r^n-1)}{r-1}\\\\
S_{30}=\frac{1(2^{30}-1)}{2-1}\\\\
S_{30}=2^{30}-1\\\\$$

$${{\mathtt{2}}}^{{\mathtt{30}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{1\,073\,741\,823}}$$  cents

So the daily payment will add up to    $10,737,418.23

The up front amount is                        $2,000,000.00

Well the daily amount is more than 5 times the up front payment.

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

I got

a=11/12 +ki

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