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scoray [572]
3 years ago
6

Algbra two, i need help asap please

Mathematics
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

(a) The zeroes of f(x) are rational numbers

(b) The smallest integer value of a is 3

Step-by-step explanation:

In the quadratic function f(x) = ax² + bx + c, we can know the number and the types of its roots using the discriminant  Δ, where Δ = b² - 4ac

  • If Δ > 0, then the zeroes are two different real numbers
  • If Δ = 0, then the zeroes are the same real number
  • If Δ < 0, then the zeroes are imaginary numbers

Note: In the first case if Δ is a perfect square number then the zeroes are rational, if not then the zeroes are irrational

(a)

∵ The equation function is f(x) = ax² + 11x + 12

∵ a = 2

∴ a = 2, b = 11 and c = 12

→ Substitute them in the rule of Δ

∵ Δ = (11)² - 4(2)(12) = 121 - 96

∴ Δ = 25

∵ Δ > 0

∴ f(x) has two different real zeroes

∵ 25 is a perfect square number

→ That means the zeroes of the function are rational numbers

∴ The zeroes of f(x) are rational numbers

(b)

∵ f(x) ha imaginary zeroes

→ By using the third rule above

∴ Δ < 0

∵ a = a, b = 11 and c = 12

→ Substitute them in the rule of Δ

∴ (11)² - 4(a)(12) < 0

∴ 121 - 48a < 0

→ Add 48a to both sides

∵ 121 - 48a + 48a < 0 + 48a

∴ 121 < 48a

→ Divide both sides by 48

∴ 2.52083 < a

∴ a > 2.52083

∵ The smallest integer greater than 2.52083 is 3

∴ The smallest integer value of a is 3

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Answer:

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Step-by-step explanation:

1. Let's review the information given to solve the question.

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