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
∵ 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
1 is the smallest possible integer.
Given that: [tex]f(x)=ax^2-2x+5[/tex].
The discreminant of this equation is:
[tex]b^2-4ac=(-2)^2-4a(5)=4-20a[/tex]
For imaginary zeros, we must get discreminant less than 0.
So,
[tex]4-20a<0\\ 20a>4\\ a>\frac{4}{20}\\ a>0.2[/tex]
Next integer of 0.2 is 1.
So 1 is the smallest possible integer.
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