For what values of a does the quadratic inequality have no solution `2ax^2+(a+3)x+a-1>0`
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) = -7a^2 + 14a + 9 < 0`
The roots of `-7a^2 + 14a + 9 = 0` are given by (using the quadratic formula)
`(-14 +- sqrt(14^2 - 4(-7)(9)))/(-14)`
`= 1 +- sqrt(448)/14 = 1 +- sqrt(32/14) = 1 +- sqrt(16/7) = 1 +- 4/sqrt(7)`
Since the determinant is a negative quadratic it is concave and so is negative below its first root and above its second root.
Therefore, the inequality has no real solutions when
`a: a in (-oo,1-4/sqrt(7))uu(1+4/sqrt(7),oo)`
You should know that the quadratic is strictly positive if the quadratic equation has no real solutions such that:
`Delta < 0`
You need to evaluate the discriminant of quadratic equation `ax^2+bx+c = 0` such that:
`Delta = b^2 - 4ac`
You need to identify `a,b,c` such that:
`a=2a , b = a+3, c= a - 1`
`Delta = (a+3)^2 - 4*2a*(a - 1)`
`Delta = a^2 + 6a + 9 - 8a^2 + 8a`
`Delta = -7a^2 + 14a + 9`
You should find the intervals for `7a^2 - 14a - 9 > 0`
You need to evaluate the solutions to -`7a^2 + 14a + 9 = 0` such that:
`-7a^2 + 14a + 9 = 0`
`7a^2- 14a- 9 = 0`
`a_(1,2) = (14+-sqrt(196 + 252))/14 => a_(1,2) = (14+-sqrt448)/14 => a_(1,2) = (14+-8sqrt7)/14 => a_(1,2) = (7+-4sqrt7)/7`
You need to notice that the expression is positive for `(-oo,(7-4sqrt7)/7)U((7+4sqrt7)/7,oo).`
Hence, the quadratic inequality has no solutions if `a in (-oo,(7-4sqrt7)/7)U((7+4sqrt7)/7,oo).`