2. Does the polynomial a4 + 4a2 + 5 = 0 have real zeroes?
Answers
Given : a⁴ + 4a² + 5 = 0
To find : Does the polynomial have real zeroes
Solution:
a⁴ + 4a² + 5 = 0
=> a²(a² + 4) = - 5
a² ≥ 0
a² + 4 ≥ 1
=> a²(a² + 4) ≥ 0
=> a²(a² + 4) + 5 ≥ 5
=> a⁴ + 4a² + 5 ≥ 5
a⁴ + 4a² + 5 ≠ 0
Hence no real zeroes
a⁴ + 4a² + 5 = 0
=> a² =( - 4 ± √16 - 20)/2
=> a² = - 2 ± i
Hence no real zero
polynomial a⁴ + 4a² + 5 = 0 does not have real zeroes
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Answer: no real zero
Step-by-step explanation: a⁴ + 4a² + 5 = 0
=> a²(a² + 4) = - 5
a² ≥ 0
a² + 4 ≥ 1
=> a²(a² + 4) ≥ 0
=> a²(a² + 4) + 5 ≥ 5
=> a⁴ + 4a² + 5 ≥ 5
a⁴ + 4a² + 5 ≠ 0
Hence no real zeroes
a⁴ + 4a² + 5 = 0
=> a² =( - 4 ± √16 - 20)/2
=> a² = - 2 ± i
Hence no real zero
polynomial a⁴ + 4a² + 5 = 0 does not have real zeroes