Prove that if coefficients of the quadratic equation ax^2 + bx + c = 0 are odd integers, then the roots of the equation cannot be rational numbers.
1The roots are rational iff b2-4ac is a perfect square
ie., b2-4ac=k2
→ b2-k2=4ac
→ (b+k)(b-k)= 4ac
4ac is even, so (b+k)(b-k) is even
→(b+k) is even and (b-k) is even
since, b is odd, k is odd....
4ac is product of odd and even
but , both (b+k) and (b-k) are evenas proved above
So, the assumption is false
Hence, b2-4ac is not a perfect square.
Therefore, the roots cannot be rational numbers
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11This had been done many times by me.....
If roots are to be rationals, then they must be factored by the mid-term factorisation....
ac is odd, so each factor k of ac is odd, whatever operation, we carry out with all kis...will lead to even integers.
But 'b' being odd, no rational root is possible.
62I think this logic you guys have given is not correct
I think you should solve it by using the difference of odd perect squares is a multiple of 8.
Incidentally i was solving this in class yesterday ;)
1That was the first thing that struck to me after seeing the question, but I din have a clear proof for it. Putting in some values made me feel that!!
After posting, I tried the proof and found it.
Difference between squares of two consecutive odd numbers is,
(2n+1)2-(2n-1)2={(2n+1)+(2n-1)}{(2n+1)-(2n-1)}
=4n(2) =8n
Therefore,
Difference between squares of any two odd numbers is, 8k
add the difference between the consecutive squares, 8(a+b+c)
Roots are rational iff b2-4ac=k2, k is odd as in #3
Now, b2-k2=8t, as proved
ie., 4ac=8t
or ac=2t, which is not possible.
Hence, roots cannot be rational.
Thank U bhaiyya[3]
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341You know Soumik, I am beginning to wonder if your posts make sense even to you. I just couldnt comprehend how you are saying something about ac being odd and hence every factor of ac is odd and then all of a sudden you bring in that b is odd. If this made sense to anyone please feel free to explain it to me
11We know that for an integer to be representable as a difference of 2 squares, it must be the product of 2 factors, both of which are either odd or even....
for 4ac, pure odd representation is not possible, but it can be represented as the product of 2 even factors viz 2a.2c and 2(2ac)....The number of ways we can represent an intger in this way - same are the number of ways we can represent it as the difference of 2 squares...
So for 4ac we have 2 representations where it is expressed as the difference of 2 squares...
(ac+1)2-(ac-1)2 is one that suffices the representation 2(2ac)
Another one is (a+c)2-(a-c)2 this one for (2a)(2c)
Above thing was general, lets now come to the problem, which unfortunately enough, admits none of the above presentations (Keep guessing why)
Hence roots are irrational.....
Hopefully this made """sense""" to atleast someone on this forum
11Sir, obviously 8 does not divide 4ac is a nice method, but what's wrong with my method? Leave out my earlier post, but what's wrong with this????? [56]
