**Question 6: Quadratic Equation Problem**Let \( p \) be a prime number and \( k \) an integer such that the quadratic equation \( x^2 + kx + p = 0 \) has two positive integer solutions. What is the value of \( k + p \)?**Options:**- ○ 1- ○ -1- ○ 0- ○ 2- ○ -2**Solution Explanation:**To solve this problem, we determine the conditions under which the quadratic equation has two positive integer solutions. The roots of the equation are solutions to:\[ x^2 + kx + p = 0 \]Using Vieta’s formulas, if the roots are \( r_1 \) and \( r_2 \), then:1. The sum of the roots \( r_1 + r_2 = -k \)2. The product of the roots \( r_1 \cdot r_2 = p \)Since \( r_1 \) and \( r_2 \) are positive integers, and \( p \) is a prime number, the only possible values for \( r_1 \) and \( r_2 \) are \( 1 \) and \( p \), giving us:\[ r_1 = 1, \, r_2 = p \quad \text{or} \quad r_1 = p, \, r_2 = 1 \]In both scenarios, \( r_1 + r_2 = 1 + p \).Thus:\[ -k = 1 + p \Rightarrow k = -(1 + p) \]To find \( k + p \):\[ k + p = -(1 + p) + p = -1 \]Hence, the correct answer is \(-1\).

Let 'a' and 'b' be the two positive integer solutions of the quadratic equation x2+kx+p=0, where p is a prime number and k, an integer. The quadratic equation having 'a' and 'b' as roots is x-ax-b=0⇒x2-a+bx+ab=0 Comparing this with x2+kx+p=0 gives: k=-a+b and p=abSince p is a prime number and p=ab, the roots must be 1 and p. Without loss of generality, take a=1 and b=p. Then k=-(1+p)=-1-p. Therefore, k + p = -1-p+p =-1 Thus, the second option is correct.

Answered: 6. Let p be a prime number and k an…

Math

Advanced Math

6. Let p be a prime number and k an integer such that x 2 + kx + p = 0 has two positive integer solutions. What is the value of k+ p? 1 O -1 O 2 O -2

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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