Solve the inequality 2x^2+4x-7<0.
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We have to solve the inequality: 2x^2 + 4x - 7 < 0
2x^2 + 4x - 7 can be written as factors of roots as
(x - x1)(x - x2), where
x1 = -4/4 + sqrt (16 + 56)/4
=> x1 = -1 + 6(sqrt 2)/4
=> x1 = -1 + (3*sqrt 2)/2
x2 = -1 - (3*sqrt 2)/2
As a product of factors we have: (x - (-1 + (3*sqrt 2)/2))(x - (-1 - (3*sqrt 2)/2)
This is negative when either of (x - (-1 + (3*sqrt 2)/2)) or (x - (-1 - (3*sqrt 2)/2) is negative.
- (x -...
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We'll have to calculate the zeroes of the quadratic to verify where the graph of the functin goes below x axis.
To determine the zeroes of quadratic, w'ell apply quadratic formula:
x1 = [-4+sqrt(16 + 56)]/4
x1 = (-4+sqrt72)/4
x1 = (-4+6sqrt2)/4
x1 = (-2+3sqrt2)/2
x1 = -1 + 3sqrt2/2
x2 = -1 - 3sqrt2/2
The graph of the quadratic function goes below x axis, over the range (-1 - 3sqrt2/2 ; -1 + 3sqrt2/2).
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