How does $x^2-8x+17=0$ have nonreal solutions? 
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.

Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17 
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?
 A: Firstly, your rearrangement should have given $x(x-8)=-17$. Secondly, $x(x-8)=c\implies x=c\lor x-8=c$ only works for $c=0$, because non-zero $c$ have other factorizations. The correct observation is that $(x-4)^2=x^2-8x+16=-1=i^2$, so $x-4=\pm i$.
A: Here, you can use the quadratic formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(17)}}{2(1)}$$
$$x=\frac{8\pm\sqrt{64-68}}{2}$$
$$x=\frac{8\pm\sqrt{-4}}{2}$$
$$x=\frac{8\pm2i}{2}$$
$$x=4\pm i $$
When you are calculating, you're trying to apply the zero product property to all numbers, which doesn't work. If the equation was
$$x(x-8)=0$$
you could determine the roots from that immediately. Since it's not $0$ (it's actually $-17$ and not $17$), you can't.
A: Your answer is not correct, please check it.
$x^{2}-8x+17=(x-4)^{2}+1=0$ by completing the square, so we have $(x-4)^{2}=-1$ and thus $x=4+i$ and $x=4-i$ are the required solutions where $i=\sqrt{-1}$.
A: Note that $\sqrt{-1}=i$$$x^2-8x+17=0 \implies x=\frac{8\pm \sqrt{64-68}}{2}=\frac{8+\sqrt{-4}}{2}=4\pm i$$
So (A) option is correct.
A: If you take $x = 17$ then $x^{2} - 8x + 17 = 289 - 56 + 17 = 362 \not=0$
Neither hold if you take $x = 25$
You can use Bhaskara's formula. $x = \frac{8+\sqrt{64 - 68}}{2} = 4 + i$ or $x = \frac{8-\sqrt{64 - 68}}{2} = 4 - i$
A: $x(x-8)=17$ is a true (but not useful) statement.
But $ab = m$ most certainly does !NOT! mean $a =m$ or $b=m$.
Consider $3\cdot 5 = 15$ where neither $3=15$ nor $5 = 15$.  Or consider $\frac 2{5}\cdot {85}2 = 17$ where neither $\frac 12$ nor $\frac {85}2$ equal $17$.
In fact $x(x-8)=17$ CAN'T mean $x =17$ unless $x-8=1$.  But $17-8 \ne 9$.  And $x-8$ CAN'T equal $17$ unless $x =1$ and $1-8 \ne 17$.
So your solution is definitely wrong.
We could true $x = \frac {17}{x-8}$ and $x-8 = \frac {17}x$ neither of which are equal to $1$ or $17$ but .... that doesn't get us anywhere.  (Also this assumes $x, x-8$ are neither $0$... which we can assume because if either were $0$ their product would have to be $0$ and not $17$.)
.....
We can't to $ab=m$ means $a=m$ or $b=m$ because that's just garbage.
But we can do $ab = 0$ means $a=0$ or $b=0$.  Why does this work for $0$ and not for $m$?
Because $0\cdot anything = 0$ always happens while $m \cdot x = m$ doesn't always happen. And because $ab=m$ where $a\ne m; b\ne m$ can happen $ab = 0$ where $a\ne 0; b\ne 0$ can NOT happen.
....
So we could solve this by factoring $x^2 -8x + 17$ to $(x+a)(x+b)$ and having $(x+a)(x+b) = 0$. so $x+a =0$ or $x+b = 0$.
The only trouble is .... we don't know how to factor $x^2 - 8x + 17$......
But we can complete the square.   If we have $x(x-8) = 17$ we can say anything about $x$ or $x-8$.  But if instead we had $(x+a)(x+a) = M$ where $x+a$ and $x+a$ are the same things and not different things. the we can conclude if $(x+a)^2 = M$ then $x+a = \pm \sqrt M$.
And we can find an $x^2 -8x + something\ that\ isn't\ 17$ is a perfect square so we can do
$x^2 - 8x + 17 =0$
$x^2 - 8x= -17$
$x^2 - 8x +something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = -17 ++something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square$
So what is $something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square$?
If $x^2 - 8x +something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = (x+ \ who \ knows)^2$ then
$x^2 - 8x +something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = x^2 + 2\ who \ knows\ x + \ who\ know^2$.
So $-8x = 2\ who\ knows\ x$ so $\ who\ knows = \frac {-8}2 = -4$.
And $something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = \ who\ know^2= (-4)^2 = 16$.
.....
SO.....
$x^2 -8x + 17 = 0$
$x^2 - 8x = -17$
$x^2 - 8x + 16 = -17 +16$
$(x-4)^2 = -1$.
Now that means $x-4 = \pm \sqrt{-1}$.

AND $\sqrt{-1} = i$!  THAT IS WHERE THE $i$ CAME FROM!

And so $x - 4 =\pm i$
And $x = 4 \pm i$.  So $x = 4+i$ or $x = 4-i$.
A: Also note:
If the solutions are $a, b$ then $(x-a)(x-b) = 0$ and $x^2 -(a+b)x + ab = 0$ and as $x^2 - 8x +17 =0$ we would have $a+b = 8$ and $ab = 17$.
Option 1:  $4 + i$ and $4- i$.  Then $(4+i) + (4-i) = (4+4) + (i-i) = 8$.  So for so good.  $(4-i)(4+i) = 16 + 4i - 4i -i^2 = 16 -i^2 = 16 -(-1) =16+1 = 17$.  That works.
Option 2: $-8, 17$.  Then $-8 + 17 =9 \ne 8$ so no good.  ANd $(-8)\cdot 17=-onehundredsomethingbig \ne 17$.  Really no good.
Option 3: $4$ and $-4$.  Then $4+ (-4) = 0\ne 8$.  No good.  And $4\times (-4) = -16 \ne 17$. No good.
Option 4: $\sqrt{17},-\sqrt{17}$.  Then $\sqrt{17}+(-\sqrt{17}) = 0\ne 8$.  And $\sqrt{17}\cdot (-\sqrt{17}) = -17\ne 17$.  So no good.
