Quadratic Inequalities

Here we consider inequalities of the form `ax^2+bx+c>0` or `ax^2+bx+c<0`, where `a!=0`.

Fact. If discriminant `D=b^2-4ac` of quadratic function `ax^2+bx+c` is negative and `a>0`, then for all x `ax^2+bx+c>0`.

Now, let's consider case when `D>=0`. To solve inequality `ax^2+bx+c>0` (or `ax^2+bx+c<0`) we need to factor `ax^2+bx+c`: `ax^2+bx+c=a(x-x_1)(x-x_2)` and then divide both sides of inequality `a(x-x_1)(x-x_2)>0` (or `a(x-x_1)(x-x_2)<0`) by a, changing sign of inequality if a<0, i.e. obtaining inequality `(x-x_1)(x-x_2)>0` (or `(x-x_1)(x-x_2)<0`). Now we need only to use fact that product of two numbers is positive (negative), if multipliers have same (different) signs.

Example 1. Solve `2x^2+5x+2>0`.

Let's find roots of `2x^2+5x+2`.

From equation `2x^2+5x+2=0` we found that `x_1=-2, x_2=-1/2`.

Thus, `2x^2+5x+2=2(x+2)(x+1/2)` and we obtain inequality `2(x+2)(x+1/2)>0`, i.e. `(x+2)(x+1/2)>0`.

This means that `x+2` and `x+1/2` should have same signs, i.e.

`{(x+2 >0),(x+1/2>0):}` or `{(x+2<0),(x+1/2<0):}`.

From first system we have that `x> -1/2`, from second system `x< -2`. Taking union of these intervals we have that solution of the initial quadratic inequality is `(-oo,-2)uu(-0.5;+oo)` (`uu` is union sign).

Example 2. Solve `5-3x>=2x^2`.

First we need to transform given inequality into standard form: `2x^2+3x-5<=0`.

Let's find roots of `2x^2+3x-5`.

From equation `2x^2+3x-5=0` we found that `x_1=-5/2, x_2=1`.

Thus, `2x^2+3x-5=2(x+5/2)(x-1)` and we obtain inequality `2(x+5/2)(x-1)<=0`, i.e. `(x+5/2)(x-1)<=0`.

This means that `x+5/2` and `x-1` should have different signs, i.e.

`{(x+5/2 <=0),(x-1>=0):}` or `{(x+5/2>=0),(x-1<=0):}`.

First system doesn't have solutions, from second system `-5/2<=x<=1`. Therefore solution of the initial quadratic inequality is `-5/2<=x<=1`.