23let the eqn hav rational roots
then D = b2-4ac = perfect square
b = odd = 2k+1
b2=(2k+1)2=odd
a=odd, c = odd , so 4 ac = 4 p ( p =odd)
so D=b2-4ac = odd - even = odd =a perfect square and an odd no = (2m+1)2
so (2m+1)2= (2k+1)2-4p
4p = (2k+1)2-(2m+1)2= (2k+2m+2)(2k-2m)=4(k+m+1)(k-m)
so p = (k+m+1)(k-m)
p = odd ,
k,m = any integers
now for any combination of k,m , if( k+m+1)=odd, then (k-m )= even, and vice versa
so odd x even = even , but p = odd
hence contradiction hence roots cant be rational
tnx sonne [6]
1wow :D amazing qwerty.. so fast ..
nishant tks for the "brain train qn of the day"
i almost got ...parr i so slow..:(
62This is a very old one.. i remember having solved this one during my preps :P
1one more exlpaination...
What we actually used to do to find the root of equation in 7th grade when we didnot have ne formulas like b2...and all??
we used to multiply a and c....and express b as summation of a.c [or simply factorize the equation].
since a and c are both odd..thier product is also odd.
Now we know that b is also odd....so how is it ever possible to have two rational numbers whose pruduct as well as sum is odd.
Hence prooved:)
62
Lokesh Verma
·2010-08-02 01:02:50
a very good proof by archana.. but works only for integers.. you have proved that there are no integral roots...
Can someone try and prove the same for rationals too..
Hint: p/q is a root... Think what happens if both are odd.. if p is even and if q is even...
1Roots will be rational only if b2-4ac is a perfect square.
Assume ( b2-4ac)=p2
p has to be odd since a,b,c are all odd
(b+p)(b-p)=4ac
b+p and b-p are both even ..so b+p = 2m and b-p = 2n
4mn=4ac
mn=ac
since ac is odd m and n both have to be odd
m = (b+p)/2 ...n = (b-p)/2
m2-n2=bp
LHS is even ..RHS is odd..this contradicts our assumption
Hence proved.
62
Lokesh Verma
·2010-08-04 23:35:14
that is great work from satya.. I also wanted a solution based on archana's logic [1]
1
SatyaPriya Ojha
·2010-08-05 01:01:04
let's assume the equation has a rational root (p/q) where p and q are reduced so that p and q don't have any common factors
ap2+bpq+cq2=0
because RHS is even
if p is even q has to be even
which contradicts the fact that p and q have no common factors
similarly if q is even p is also even
hence
if p is odd q has to be odd and vice versa
but if p and q are both odd
ap2+bpq+cq2 is also odd (hence can't be 0)
Proved :)
62
Lokesh Verma
·2010-08-05 10:51:06
Perfectly done my friend :) [1]
39once again a old one recasted
i already did this here:
http://www.targetiit.com/iit-jee-forum/posts/quad-14741.html