Given that p and q are prime, Find the solutions of px^2 - qx + q = 0 .
@ScareCrow28 said:Given that n is a natural number, when is n^4 + 4 prime?
Solution:
n^4 +4 = n^4 + 4n^2 +4 - 4n^2 = (n^2+2n+2)(n^2-2n+2) = Prime
=> One of them is = 1
n^2 + 2n + 2 can't be 1
So from other, n =1 is only solution
@ScareCrow28 said:Given that p and q are prime, Find the solutions of px^2 - qx + q = 0 .
no solutions?? 
@ScareCrow28 said:Given that p and q are prime, Find the solutions of px^2 - qx + q = 0 .
Sum of roots = product of the roots. And both are prime.
Hence p = q = 2
@sbharadwaj said:Sum of roots = product of the roots. And both are prime. Hence p = q = 2
No rational solution exist..
@sbharadwaj said:Sum of roots = product of the roots. And both are prime. Hence p = q = 2
x^2 - x + 1 =0 yields imaginary roots...
@Logrhythm said:x^2 - x + 1 =0 yields imaginary roots...
lekin wo sum =product wali baat bhi sahi hai na..
fir kuchh to gadbad hai na............
@Logrhythm said:x^2 - x + 1 =0 yields imaginary roots...
Oops.. 2 and 2 should be the roots ( not p and q) , in which case p will be 1
Hence no solutions.!
PS: thanks for correcting :)
@ScareCrow28 said:Given that p and q are prime, Find the solutions of px^2 - qx + q = 0 .
no solution ??
@ScareCrow28 said:how?
bhai meine toh hit and trial karke dekha tha 3 4 number se....nahi aya toh bol dia no solution....hehe...algebra aese hi hoti hai mujhse...
waise aese kuch kar sakte hai kya...
px^2 - qx + q = 0
D = q^2 - 4pq = q(q-4p)
for rt(D) to be an integer, q and (q-4p) must be a perfect square which isn't possible so no solutions again....
@Logrhythm said:bhai meine toh hit and trial karke dekha tha 3 4 number se....nahi aya toh bol dia no solution....hehe...algebra aese hi hoti hai mujhse...waise aese kuch kar sakte hai kya... px^2 - qx + q = 0D = q^2 - 4pq = q(q-4p) for rt(D) to be an integer, q and (q-4p) must be a perfect square which isn't possible so no solutions again....
Fair enough! 😃 OR you could have dine like this:
D = q^2 -4pq = (2p-q)^2 - (2p)^2 which is always
Hence D
@ScareCrow28 said:Fair enough! OR you could have dine like this: D = q^2 -4pq = (2p-q)^2 - (2p)^2 which is always Hence D
p=2 and q=11/13/17 etc
inn sab ke liye nahi true hoga bhai, i started off with this intent only and then moved to what i wrote...
@Logrhythm said:p=2 and q=11/13/17 etcinn sab ke liye nahi true hoga bhai, i started off with this intent only and then moved to what i wrote...
Ohhh yes.. 
True.. Then what you wrote is cool ..
True.. Then what you wrote is cool .. @ScareCrow28 said:Given that p and q are prime, Find the solutions of px^2 - qx + q = 0 .
@Logrhythm bhai iska answer straight away conclude kar sakte the rt(q^2-4pq) will never yield a integer value
Good Morning Guys,
sab kahan gayab ho?
When a two digit number is added to another two digit number having the same digits in reverse order, the sum is a perfect square.How many such 2 digit numbers are there?
4
6
8
10
4
6
8
10
@saurav205 said:Good Morning Guys,sab kahan gayab ho?When a two digit number is added to another two digit number having the same digits in reverse order, the sum is a perfect square.How many such 2 digit numbers are there?46810
10a + b + 10b + a = k^2
11*(a + b) = k^2
a + b = 11
=> 4 solutions
so , in total 8 such numbers
@saurav205 said:Good Morning Guys,sab kahan gayab ho?When a two digit number is added to another two digit number having the same digits in reverse order, the sum is a perfect square.How many such 2 digit numbers are there?46810
ab + ba = (10a + b) + (10b + a) = 11(a + b) is a perfect square.
Clearly (a + b) should be 11 times perfect square..(think why??)
Now as a and b are single digit positive integers, we have 1
So Only multiple of 11 in the above range is 11 which is 11*1^2.
Thus (a + b) = 11 and possible such numbers are (a, b) = (2, 9), (3, 8), ..., (8, 3), (9, 2) i.e. 8 solutions..
Team BV
Clearly (a + b) should be 11 times perfect square..(think why??)
Now as a and b are single digit positive integers, we have 1
So Only multiple of 11 in the above range is 11 which is 11*1^2.
Thus (a + b) = 11 and possible such numbers are (a, b) = (2, 9), (3, 8), ..., (8, 3), (9, 2) i.e. 8 solutions..
Team BV
@bodhi_vriksha said:ab + ba = (10a + b) + (10b + a) = 11(a + b) is a perfect square.Clearly (a + b) should be 11 times perfect square..(think why??)Now as a and b are single digit positive integers, we have 1 So Only multiple of 11 in the above range is 11 which is 11*1^2.Thus (a + b) = 11 and possible such numbers are (a, b) = (2, 9), (3, 8), ..., (8, 3), (9, 2) i.e. 8 solutions..Team BV
The reason is 11 is a prime number..so to get a perfect square the other number should also be a 11...
11 times perfect square will give you 2 11s and a perfect square...so when you take the square root you get one 11 and sq rt of the (perfect square) , which in this case comes out to be 1..
P.S. : BTW what is Team BV?? Bodhi Vriksha??