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 solutions. The value of (k+p) equals

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 solutions. The value of (k+p) equals (A) -1 (B) 0 (C) 3 (D) 6

Tardigrade Admin · Accepted Answer

Let r 1 and r 2 are two integral solutions r1+r2=-k text and r1 r2=p since p is prime hence either r 1=1 or r 2=1 let r1=1 ; r2=p 1+p=-k ⇒ k+p=-1

Q. Let $k$ be an integer and $p$ is a prime such that the quadratic equation $x^{2} + k x + p = 0$ has two distinct positive integer solutions. The value of $(k + p)$ equals

-1

0

3

6

Let $r_{1}$ and $r_{2}$ are two integral solutions
$r_{1} + r_{2} = - k and r_{1} r_{2} = p$
since $p$ is prime hence either $r_{1} = 1$ or $r_{2} = 1$
let $r_{1} = 1; r_{2} = p$
$1 + p = - k$
$\Rightarrow k + p = - 1$

Q. Let k be an integer and p is a prime such that the quadratic equation x2+kx+p=0 has two distinct positive integer solutions. The value of (k+p) equals

-1

0

3

6

Let r1​ and r2​ are two integral solutions r1​+r2​=−k and r1​r2​=p since p is prime hence either r1​=1 or r2​=1 let r1​=1;r2​=p 1+p=−k ⇒k+p=−1

Q. Let $k$ be an integer and $p$ is a prime such that the quadratic equation $x^{2} + k x + p = 0$ has two distinct positive integer solutions. The value of $(k + p)$ equals

Let $r_{1}$ and $r_{2}$ are two integral solutions
$r_{1} + r_{2} = - k and r_{1} r_{2} = p$
since $p$ is prime hence either $r_{1} = 1$ or $r_{2} = 1$
let $r_{1} = 1; r_{2} = p$
$1 + p = - k$
$\Rightarrow k + p = - 1$