For what values of a does the quadratic inequality have no solution `2ax^2+(a+3)x+a-1>0`
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If the inequality `2ax^2 + (a+3)x + a -1 > 0` is to have no solution
this implies that the quadratic `2ax^2 + (a+3)x + a - 1` is strictly negative
ie that the quadratic `-2ax^2 - (a+3)x - a + 1` is strictly positive.
A quadratic is strictly positive when it has no real roots, that is, when it's determinant is negative.
The determinant is given by `B^2 - 4AC` where `A ` is the coefficient of `x^2`, `B` is the coefficient of `x` and `C` is the zeroth order term.
Here, `A = -2a`, `B = -(a +3)` and `C = 1 -a`
Therefore the determinant is given by `B^2 - 4AC = (a+3)^2 - 4(-2a)(1-a)`
This is negative if `a^2 + 6a + 9 - 4(-2a + 2a^2) =...
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