Determine the smallest integer value of a for which f (x) has imaginary zeroes. Show how you found

this answer.

The question is incomplete, the complete question is

Determine the smallest integer value of a for which f (x) = ax² - 2x + 5 has imaginary zeroes. Show how you found this answer

Answer:

The smallest integer value of a is 1

Step-by-step explanation:

To find the zeroes of a function equate it by 0, then find the values of x which are the zeroes of the function

To find the types of the roots (zeroes) of a function f (x) = ax² + bx + c use the discriminant of the function b² - 4ac

If b² - 4ac > 0, then the function has two different real roots If b² - 4ac = 0, then the function has one real root If b² - 4ac < 0, then the function has no real roots (imaginary roots)

∵ f (x) = ax² - 2x = 5

- To find its zeroes equate f (x) by 0

∴ ax² - 2x + 5 = 0

∵ f (x) has imaginary zeroes

- That means the discriminant is less then zero

∴ b² - 4ac < 0

∵ a = a, b = - 2 and c = 5

- Substitute them in the inequality above

∴ (-2) ² - 4 (a) (5) < 0

∴ 4 - 20 a < 0

- Add 20 a to both sides

∴ 4 < 20 a

- Divide both sides by 20

∴ 0.2 < a

- That means a is greater than 0.2

∴ a > 0.2

∵ You must to find the smallest integer value of a

- The first integer greater than 0.2 is 1

∴ a = 1

The smallest integer value of a is 1