Given that the polynomial x^2-kx+160 has only positive integer roots, find the average of all distinct possibilities for k.
+171 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
+15000 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 \(\}\)
\(x=-\frac{(-k)}{2} \pm \sqrt{(\frac{k}{2})^2-160}\\ x1=\frac{28}{2} \pm \sqrt{(\frac{28}{2})^2-160}=14 \pm6 \\ \color{blue} x_1=8\\ \color{blue} x_2=20\\ x_3=\frac{44}{2} \pm \sqrt{(\frac{44}{2})^2-160}=22\pm 18\\ \color{blue} x_3=4\)
\(x_4=40\)
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