Cho 2x2 + 5x + 4 = 0
=> 2(x2 +\(\dfrac{5}{2}\)x + 2 ) = 0
=> x2 +\(\dfrac{5}{4}\)x + \(\dfrac{5}{4}\)x + \(\dfrac{25}{16}\) + \(\dfrac{7}{16}\) = 0
=> x(x + \(\dfrac{5}{4}\)) + \(\dfrac{5}{4}\)(x + \(\dfrac{5}{4}\)) = \(\dfrac{-7}{16}\)
=> (x + \(\dfrac{5}{4}\))(x + \(\dfrac{5}{4}\)) = \(\dfrac{-7}{16}\)
=> (x + \(\dfrac{5}{4}\))2 = \(\dfrac{-7}{16}\)
Ta có : (x + \(\dfrac{5}{4}\))2 \(\ge\) 0 \(\forall\)x
Mà \(\dfrac{-7}{16}\) < 0
Nên đa thức f(x) không có nhgieemj
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