Show that if #a, b, and c# are all odd integers then the quadratic #ax^2 + bx + c = 0# has no solution in the set of rational numbers?
2 Answers
The discriminant of a quadratic formula is
For a number to be odd,
Using this, we get:
By expanding, we get:
Proof:
Also,
As you can see, there are two spaces missing.
Though, if the squares of two odd numbers are multiplied together, it still gives a square number,
Also, if we take
No matter what integer you use for
See explanation...
Explanation:
If
If
#ax^2+bx+c = (px+q)(rx+s)#
#color(white)(ax^2+bx+c) = prx^2+(ps+qr)x+qs#
Equating coefficients, we have:
#{ (a = pr), (b=ps+qr), (c=qs) :}#
Since
Then:
#b = ps+qr#
is the sum of two odd integers and therefore even.
This contradicts the assumption that