1. ## Roots of equation

I've gone through my workings twice and re-did, but somehow I keep getting an irregularity.

Originally Posted by Question
The roots of the equation $x^2 - (q - 5)x = \frac{q}{2}$ are $\alpha$ and $\alpha+5$. Find the possible values of q.
I did it like this.

Since the roots are $\alpha$ and $\alpha + 5$, I substituted them into the equation.

$\alpha^2 - (q - 5)\alpha = (\alpha + 5)^2 - (q - 5)(\alpha + 5)$
$\alpha^2 - (q - 5)\alpha = \alpha^2 + 10\alpha + 25 - (q\alpha - 5\alpha + 5q - 25)$
$\alpha^2 - (q - 5)\alpha = \alpha^2 | 10\alpha + 25 - q\alpha + b\alpha - 5q + 25)$
[tex]\alpha^2 - (q - 5)\alpha = \alpha^2 + (15 - q)\alpha + (50 - 5q)[tex]

Therefore $-(q - 5)\alpha = (15 - q)\alpha$, $50 - 5q = 0$
$5 - q = 15 - q$, $q = 10$

But 5-q = 15-q doesn't even make any sense.

2. Originally Posted by fterh
I've gone through my workings twice and re-did, but somehow I keep getting an irregularity.

I did it like this.

Since the roots are $\alpha$ and $\alpha + 5$, I substituted them into the equation.

$\alpha^2 - (q - 5)\alpha = (\alpha + 5)^2 - (q - 5)(\alpha + 5)$
$\alpha^2 - (q - 5)\alpha = \alpha^2 + 10\alpha + 25 - (q\alpha - 5\alpha + 5q - 25)$
$\alpha^2 - (q - 5)\alpha = \alpha^2 | 10\alpha + 25 - q\alpha + b\alpha - 5q + 25)$
[tex]\alpha^2 - (q - 5)\alpha = \alpha^2 + (15 - q)\alpha + (50 - 5q)[tex]

Therefore $-(q - 5)\alpha = (15 - q)\alpha$, $50 - 5q = 0$
$5 - q = 15 - q$, $q = 10$

But 5-q = 15-q doesn't even make any sense.
There are some typos in your work so I don't know what you meant to write. But I think starting with

$\alpha^2 - (q - 5)\alpha = (\alpha + 5)^2 - (q - 5)(\alpha + 5)$

while valid results in a loss of information, because the $c$ term in $ax^2+bx+c=0$ gets canceled out. When I worked out everything according to your first step, I got $q=2\alpha+10$, which is a valid restriction on $q$ but is not the solution set. (Because not all values of $\alpha$ are possible.)

To solve, I did:

$\displaystyle x^2-(q-5)x-\frac{q}{2}=0$

$\displaystyle x=\frac{q-5\pm\sqrt{(q-5)^2-4(-\frac{q}{2})}}{2}$

$\dots$

$\displaystyle \frac{q-5-\sqrt{q^2-8q+25}}{2}+5 = \displaystyle \frac{q-5+\sqrt{q^2-8q+25}}{2}$

$\dots$

$10=2\sqrt{q^2-8q+25}$

$\dots$

$q\in\{0,8\}$

And also, the question doesn't ask it, but $\alpha\in\{-5,-1\}$

3. Sorry but can you highlight to me where are the typos? I looked through it and it seems to be in order except for a trailing / at the 3rd last statement.

So you mean that my method won't work?

4. Originally Posted by fterh
Sorry but can you highlight to me where are the typos? I looked through it and it seems to be in order except for a trailing / at the 3rd last statement.
Unfortunately the \color{} command doesn't work like it did in the old forum, so I can't highlight directly, but in the line you wrote

$\alpha^2 - (q - 5)\alpha = \alpha^2 | 10\alpha + 25 - q\alpha + b\alpha - 5q + 25)$

there is a misplaced "|" symbol and also the letter b out of the blue, and a stranded end parenthesis at the very end, then the next line isn't rendered in LaTeX due to the missing "/" .... well I can work out what you mean with some effort, but really, why should I put in all the effort to read what you didn't put much effort into writing/editing?

So I can see you go from

$\alpha^2 - (q - 5)\alpha = \alpha^2 + (15 - q)\alpha + (50 - 5q)$

to

$- (q - 5) = 15 - q\quad\text{AND}\quad 0 = 50 - 5q$

This doesn't work because consider a different example:

$10x = 5x + 12$

$10 = 5\quad\text{AND} \quad0 = 12$

So you got confused with this kind of situation:

If $a_1x^2 + b_1x + c_1$ and $a_2x^2 + b_2x + c_2$ are the same polynomial, then $a_1 = a_2$, $b_1 = b_2$, and $c_1 = c_2$.

Originally Posted by fterh
So you mean that my method won't work?
Well, like I said, using your method results in $q=2\alpha+10$. You would then have to go back and substitute

$\displaystyle x-(q-5)x=\frac{q}{2}$

$\displaystyle \alpha-((2\alpha+10)-5)\alpha=\frac{2\alpha+10}{2}$

and solve for $\alpha$, etc. Seems like more work, overall.

5. I see, thanks for your help

Really sorry about the typos, I must have been blind