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k is a prime number and f(x) = k has no solutions.
Find the possible values of k.​

Sagot :

agumbs0002

Answer:

Suppose x∈R is such that x+20x=p is a prime. Then x2−px+20=0 , and this equation has a solution x∈R if and only if p2–80≥0 . So the primes which admit no real solution are those that satisfy p<80−−√ , or 2 , 3 , 5 , and 7 . ■

Step-by-step explanation:

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The primes which accept no real solution exist as those that satisfy

p<[tex]\sqrt{80}[/tex], or 2, 3, 5, and 7.

What is a prime number?

A prime number exists as a real number greater than 1 whose only factors that exist are 1 and itself. A factor exists as an entire number that can be divided evenly into another number. The first prime numbers exist 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. Numbers that contain better than two factors exist named composite numbers.

If the function f(x) = (20/x) + x = k. Suppose x∈R exists such that x+20x=p exists a prime. Then [tex]$x^2-px+20=0[/tex], and this equation contains a solution x∈R if and only if [tex]$p^2-80[/tex] ≥ 0. So the primes which accept no real solution exist as those that satisfy p < [tex]\sqrt{80}[/tex], or 2, 3, 5, and 7.

To learn more about prime number refer to:

https://brainly.com/question/145452

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