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Given that the polynomial x^2-kx+160 has only positive integer roots, find the average of all distinct possibilities for k.

 Aug 9, 2021
 #1
avatar+171 
+6

first of all we are given that the roots $\in \mathbb{Z}^{++}$. going easy on this, by using vietas formula we know that the sum of the roots of that polynomial is $\frac{16}{a}$, and the product of them is $\frac{-(-k)}{a}$ where $a=1$ right?


so, this polynomial is such that it has got these three solutions: $(16,1) \ \wedge\ (8,2) \ \wedge\ (4,4)$ , therefore the sum of all those would give us $k \{ 17 \ \wedge\ 10 \ \wedge\ 8 \}$

 

adding them alltogether we get $35$ which when divided by three gives us $11. \overline{6}  $ or more correctly $11\frac{2}{3}$rds

 Aug 9, 2021
 #2
avatar+14903 
+2

Given that the polynomial x^2-kx+160 has only positive integer roots, find the average of all distinct possibilities for k.

 

Hello Guest!

 

\(x^2-kx+160\\ x=-\frac{(-k)}{2} \pm \sqrt{(\frac{k}{2})^2-160}\\ [(\frac{k}{2})^2-160]\in\{Square\ numbers \}\\ [(\frac{k}{2})^2-160]\in\{  1 , 4,  9, 16,  25,  36 ,  49 , 64 , 81\\ , 100,  121 , 144 , 169, 196 , 225 ,  256 , 289 , 324 ,  361 , 400 , 441 , 484 , 529 ,  576 , 625 \}\)

\([(\frac{k}{2})^2-160]\in \{\)9,    36,   65,   96,   129, 164, 201,  240, 281,  324, 369,  416,  465\(\}\)  

      \(k\in\{\)              \(2\sqrt{196},\)                                                       \(2\sqrt{484}\ \}\)

 

      \(\color{blue} k\in\{\)                    28,                                                                44   \(\}\)

 

  \(x=-\frac{(-k)}{2} \pm \sqrt{(\frac{k}{2})^2-160}\\ x1=\frac{28}{2} \pm \sqrt{(\frac{28}{2})^2-160}=14 \pm6 \\ \color{blue} x_1=8\\ \color{blue} x_2=20\\ x_3=\frac{44}{2} \pm \sqrt{(\frac{44}{2})^2-160}=22\pm 18\\ \color{blue} x_3=4\)

  \(x_4=40\)

laugh  !

 Aug 9, 2021
edited by asinus  Aug 10, 2021
edited by asinus  Aug 10, 2021
edited by asinus  Aug 10, 2021
edited by asinus  Aug 10, 2021

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