Question

let K be an integer and p is a prime such that the quadratic equation x^2+Kx+P=0 has two distinct positive integer solution.The value fo K-P equals to

Answer 1

@AllTehMaffs pls...hlp

Answer 2

i also did upto dat and afterward my brain stopped wrking

Answer 3

\[K-P > 2 \sqrt P - P\] o_0

Answer 4

but P is a prime, so there's no such integer as sqrt P

Answer 5

wt i m notr able to undrstand

Answer 6

hmmm

\[K^2 -4P > 0\]
\[(K+2P)(K-2P)>0\]
so
\[K>-2P\]
\[K>2P\]
?

Answer 7

this isn't helpful I think :P

Answer 8

if P is prime then only 1 divides it, so \sqrt P can't be an integer. But I don't think that's helpful in this problem.

Answer 9

the answer is 1
bt how

Answer 10

Well the only solution that satisfies this inequailty
\[K-P > 2\sqrt P - P\]
is P = 1, then the answer on the right hand side jas to be 1

Answer 11

hw u did that

Answer 12

given that k is an integer and p is a prime so......both the roots will be integers coz considering the case of fractional can satisfy the condn of K but not of P.....so we know that roots are integers.... then multiplying two integers can be only prime when one of them is 1 otherwise its not possible, so we have one root as 1

Answer 13

then K= (1+a) and P = 1*a...... K-P = 1

Answer 14

that's certainly gud .......

Answer 15

the way I did it was
\[K^2 - 4P > 0\] because of the quadratic formula, then
\[K^2 > 4P\]
\[K>2\sqrt P\]
\[K-P > 2\sqrt P - P\]
The only prime that has an integer square root is 1, , so P=1
\[K-P > 2-1\]
K-P > 1

But it's still an inequality; I think div.u might have a more solid solution

Answer 16

Am I interrupting?

Answer 17

no bt i got da answer @divu.mkr explained me

Answer 18

Okay then, good luck.