Question

x^2-(a-3)x+a-2=0 has two integral solutions. List all possible 'a's.

Answer 1

are you supposed to use integrals or just solve that equation?

Answer 2

If b^2-4ac < 0, no solutions exist
If b^2-4ac = 0, one solution exists.
If b^2-4ac > 0, two solutions exist.

Answer 3

okay so in other words, we have to study the behaviour of "a".
So let's write it on those terms:
\[-(a-3)^{2}-4(a-2)\]
let's expand that binome and apply distributive on the other hand:
\[-a ^{2}+6a-9-4a+8\]
let's operate similar terms:
\[-a ^{2}+2a-1\]
now I want to know what value of "a" makes that 0 so:
\[-a ^{2}+2a-1=0\]
and we get that a=1.
so we found a value of a that determines a horizon in the discriminant part, so we can say:

when a<1
\[-(a-3)^{2}-4(a-2) <0 \] there are no solutions for the equation

when a=1
\[-(a-3)^{2}-4(a-2) = 0\] there is only one solution for the equation

when a>1
\[-(a-3)^{2}-4(a-2)>0\] there are two solutions for the equation.

we the list of all the "a" values that give us 2 solutions are [1,+inf)

Answer 4

*

Answer 5

oooh, why u stop? :( That was very interesting.

Answer 6

we want t come up with 2 integral solutions, so discriminant of quadratic must be a positive square, in mathematical words\[\Delta=(a-3)^2-4(a-2)=a^2-10a+17=m^2\]

Answer 7

\(m\) is a positive integer (why?) and\[a^2-10a+17=(a-5)^2-8=m^2\]\[a=5\pm\sqrt{m^2+8}\]this is necessary but not sufficient, let's check the solutions\[x_1,x_2=\frac{a-3\pm\sqrt{a^2-10a+17}}{2}=\frac{2\pm\sqrt{m^2+8}\pm m}{2}\]

Answer 8

now \(\pm m\pm\sqrt{m^2+8}\) must be an even integer (why?) it's possible only if \(m=1\) (why?)

Answer 9

finally \(a=8,2\) :)

Answer 10

hohohoho, I'll take note of that method, looks interesting to do some profound research.