what values of R (all real numbers) are solutions to the quadratic inequality `2ax^2 + (a-3)x + a -1 <0`
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First solve the quadratic using the quadratic formula
`x = (-B +- sqrt(B^2 - 4AC))/(2A)`
where `A = 2a`, `B = a-3` and `C = a -1`
This gives solutions
`x = (3-a +- sqrt((a-3)^2 - 4(2a)(a-1)))/(4a)`
` = (3-a +- sqrt(a^2 - 6a + 9 - 8a^2 + 8a))/(4a)`
`= (3-a +- sqrt(-7a^2 + 2a + 9))/(4a)`
Now, the determinant `-7a^2 + 2a + 9` factorizes to `(9-7a)(a+1)`
and because `-7a + 2a + 9` is concave then it is positive between it's two roots
`a = -1` and `a = 9/7`
When the determinant is positive, there are at least two real roots, so
the equation has real solutions when `a: a in [-1,9/7]` answer
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