1. ## two quadratic equation problems

2. Originally Posted by afeasfaerw23231233
$\displaystyle \left\{ {\begin{array}{ll} xy&= 50 \\ x^{\log y}&= 5 \end{array}} \right.$
$\displaystyle xy = 50 \implies x = \frac{50}{y} \implies \log x = \log 5 + 1 - \log y$

thus, $\displaystyle x^{\log y} = 5 \implies \log y \log x = \log 5$

therefore $\displaystyle \log y (\log 5 + 1 - \log y) = (\log 5 + 1) \log y - (\log y)^2 = \log 5$

$\displaystyle \implies (\log y)^2 - (\log 5 + 1)\log y + \log 5 = 0$

setting $\displaystyle u = \log y$, then you are just solving a quadratic equation in 1 variable..

3. thanks but how to do question 2 [ii]? how to show k = p-1

4. My solution is a little involved
I call $\displaystyle \alpha$ as 'a' and $\displaystyle \beta$ as 'b' to avoid TeX.

First of all ,if a and b are integers, so are a-1 and b-1.
By part {i}, they satisfy $\displaystyle x^2+(k+2)x+ p = 0$'

So if a-1 and b-1 satisfy $\displaystyle x^2+(k+2)x+ p = 0$
then (a-1)(b-1) = p.

Since p is prime,the only possible solutions are
(a,b) = (2,p+1),(p+1,2),(0,-p-1),(-p-1,0)

We first claim that the first two solutions are invalid.

If either of a or b were 2, then they satisfy x^2 + kx + (p-k-1) = 0
that means $\displaystyle 4+2k+p-k-1= 0 \Rightarrow 3+p+k = 0$
But this is not possible since p>0 and k>0 .

So look at the remaining solutions,

Now that means either of a or b is 0.
Substitute in x^2 + kx + (p-k-1) = 0
And get k = p-1

All done again