The number of zeroes that polynomial f(x) = (x – 2)² + 4 can have is: *
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Answers
Given : f(x) = (x – 2)² + 4
To Find : The number of zeroes that polynomial f(x) = (x – 2)² + 4 can have
Solution:
f(x) = (x - 2)² + 4
= x² - 4x + 4 + 4
= x² - 4x + 8
Comparing with ax² + bx + c
a = 1 , b = -4 , c = 8
D = b² - 4ac
=> D = (-4)² - 4(1)(8)
= 16 - 32
= -16
D < 0
Hence no real root
So number of zeroes that polynomial f(x) = (x – 2)² + 4 can have is 0
other method :
f(x) = (x – 2)² + 4
least value of (x – 2)² is 0
Hence least value of f(x) is 0 + 4 = 4
Hence f(x) ≠ 0
so no zeroes
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