Problem 1
Let A be the set of numbers a, for which the equation [tex](a+10)(a+8)(a+6)(a+4)(a+2)a(a-2)(a-4)(a-6)(a-8)(a-10)x=a+10[/tex] has no solutions for x. Determine the product of all elements of A.
Problem 2
Let [tex]A (a_1,a_2,\dots ,a_n)[/tex] be the set of natural numbers a, which satisfy the condition [tex]3<2a<15[/tex]. Let [tex]x_1,x_2,\dots,x_n[/tex] be the solutions to the equation [tex](a^2-1)x=a^2+2a+1[/tex]. Find the product [tex]x_1 x_2 \dots x_n[/tex]
Solution:
We'll first determine the set [tex]A[/tex].
[tex]3<2a<15[/tex] is equivalent to
[tex]\begin{array}{|l}2a>3 \\ 2a<15\end{array}[/tex]
[tex]\begin{array}{|l}a>\frac{3}{2}=1\frac{1}{2} \\ a < \frac{15}{2}=7\frac{1}{2} \end{array}[/tex]
So, the numbers [tex]a_1,a_2,\dots,a_n[/tex] are all the integers (all of them natural numbers) between [tex]1\frac{1}{2}[/tex] and [tex]7 \frac{1}{2}[/tex]. These numbers are [tex]2,3,4,5,6,7[/tex] and respectively [tex]a_1=2,a_2=3,a_3=4,a_4=5,a_5=6,a_6=7[/tex].
We'll also find an explicit formula for the solution of the equation.
[tex](a-1)(a+1)x=(a+1)^2[/tex]. Obviously [tex]a \ne -1[/tex], we divide by [tex]a+1[/tex] and get
[tex](a-1)x=a+1[/tex]. Since each [tex]a \ge 2[/tex], [tex]a-1 \ne 0[/tex] and the solutions are
[tex]x=\frac{a+1}{a-1}[/tex]. We substitute with each a:
[tex]x_1=\frac{a_1+1}{a_1-1}=\frac{2+1}{2-1}=3[/tex]
[tex]x_2=\frac{a_2+1}{a_2-1}=\frac{3+1}{3-1}=\frac{4}{2}=2[/tex]
[tex]x_3=\frac{a_3+1}{a_3-1}=\frac{4+1}{4-1}=\frac{5}{3}[/tex]
[tex]x_4=\frac{a_4+1}{a_4-1}=\frac{5+1}{5-1}=\frac{6}{4}=\frac{3}{2}[/tex]
[tex]x_5=\frac{a_5+1}{a_5-1}=\frac{6+1}{6-1}=\frac{7}{5}[/tex]
[tex]x_6=\frac{a_6+1}{a_6-1}=\frac{7+1}{7-1}=\frac{8}{6}=\frac{4}{3}[/tex]. We multiply them:
[tex]x_1x_2x_3x_4x_5x_6=3\cdot2\cdot\frac{5}{3}\cdot\frac{3}{2}\cdot\frac{7}{5}\cdot\frac{4}{3}=\frac{3\cdot2\cdot5\cdot3\cdot7\cdot4}{3\cdot2\cdot5\cdot3}=7\cdot4=28[/tex]
Problem 3
Find the value of a, for which the equation
[tex](a^2-7a+10)x=a^2-8a+15[/tex] has all [tex]x \in R[/tex] as solutions.