Quadratic expression concept on inequality - II

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We have learnt from previous post that,

f(x) = a{x^2} + bx + c, a \ne 0


and we have seen that there are two roots of f(x)
{x_1} = \dfrac{{ - b - \sqrt D }}{{2a}},\,\,{x_2} = \dfrac{{ - b + \sqrt D }}{{2a}}
*Roots of any equation are those value of x for which f(x) = 0 or y = 0
and on x-axis the value of y = 0... in fact equation of x -axis is y = 0



        Here curve cuts the x-axis on two points {x_1} and {x_2} on these points the value of y = 0 or f(x) = 0,
therefore roots are {x_1} and {x_2}.

* f(x) = 0 means two curves y = f(x) and y = 0 , and we have to find that on which values of x these two curves are equals , which also means on which points these curves cuts or intersects with eachother.
we already know that for y = {x^2} , the vertex of the curve is (0,0) i.e. on origin

 for general equation, f(x) = a{x^2} + bx + c, a \ne 0
or, f(x) = a{\left( {x + \frac{b}{{2a}}} \right)^2} - \frac{D}{{4a}} (from previous post)

 if we shift the origin to \left( { - \frac{b}{{2a}}, - \frac{D}{{4a}}} \right), the equation becomes Y = a{X^2},
 therefore vertex of general quadratic equation is \left( { - \frac{b}{{2a}}, - \frac{D}{{4a}}} \right)

Next, 

 f(x) = a{x^2} + bx + c, a \ne 0
we have seen that there are two roots of f(x)
{x_1} = \dfrac{{ - b - \sqrt D }}{{2a}},\,\,{x_2} = \dfrac{{ - b + \sqrt D }}{{2a}}

but we know that square root of negative number does not exist i.e imaginary roots
means if D < 0 we have imaginary roots means curve never cuts the x-axis i.e. curve never gets the value '0'

 and for D = 0 both roots have same value i.e. {x_1} = {x_2} = \frac{{ - b}}{{2a}} (put D = 0)
means vertex touch on x-axis recall x = - \frac{b}{{2a}} for vertex and y = - \frac{D}{{2a}} if D = 0 then y = 0 

 Now, understand and learn the following concepts,

 1. a > 0 and D < 0

  clearly, quadratic expression will always be greater than ZERO.. means y > 0 always......
i.e.f(x) > 0

     so for any quadratic expression if a > 0 and D < 0 .... the quadratic expression is always greater than zero.

 e.g. 2{x^2} + 2x + 5,
a = 2 > 0, D = {2^2} - 4.2.5 = - 36 < 0
\Rightarrow (2{x^2} + 2x + 5) > 0

 2. a < 0 and D < 0

               

 clearly, quadratic expression will always be less than zero.. means y < 0 always......
i.e.f(x) < 0

     so for any quadratic expression if a < 0 and D < 0 .... the quadratic expression is always less than zero.

 e.g.  - 3x + 2x - 5, a = - 3 < 0, D = 4 - 60 = - 56 < 0
\Rightarrow ( - 3x + 2x - 5) < 0

             

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