0≤x²+3x-4<1
1个回答

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0 ≤ x² + 3x -4 < 1

分两部分算:

(1)

x² + 3x -4 ≥ 0;

(x - 1)(x + 4) ≥ 0 ;

x≥1 或 x≤-4

(2)

x² + 3x -4 < 1;

x² + 3x -5 < 0;

用求根公式算:

△ = b² -4ac = 3² - 4 * 1 * (-5) = 29

x1 = (-b + √△)/2a = (√29 - 3)/2

x2 = (-b - √△)/2a = (-√29 - 3)/2

x2 < x < x1

(-√29 - 3)/2 < x < (√29 - 3)/2

结合(1)(2):

1 ≤ x < (√29 - 3)/2 或 (-√29 - 3)/2 < x ≤ -4