Property 3
In solving an inequality system always positive expressions are cancelled.
- \(|f(x)|\ge0\) — Expression of absolute value
- \(n\in\mathbb{N},\ [f(x)]^{2n}\ge0\) — Squared expression
- \(a^x>0,\ a>0\) — Exponential function
- \(f(x)=ax^2+bx+c<0,\quad\Delta<0,\quad a>0\)
Quadratic equation without root: If “a” is a negative number it will be cancelled and only the direction of the sign will change.
- \(n\in\mathbb{N}\)\(f(x)>0\Rightarrow[f(x)]^{2n-1}>0\)\(f(x)<0\Rightarrow[f(x)]^{2n-1}<0\)
As the signs of \(f(x)\) and \([f(x)]^{2n-1}\) are the same the power of the exponential expression will be cancelled.
If expressions can be simplified then they are simplified but only the roots will be analysed.
Note: The roots of the cancelled expressions should be analyzed.
Example
\(\samefrac{|x-3|\cdot3^x\cdot(x-2)\cdot(x+7)^4\cdot(x-9)}{(x^2+x+7)\cdot(x+5)^{99}\cdot(x-9)}\ge0\)
\(\Rightarrow S.S.=?\)
Answer
- \(|x-3|\ge0\)
canceled
\(x-3=0\Rightarrow x=3\quad\text{(could be)}\) - \(3^x>0\)
Exponential function are canceled. No root.
- \((x+7)^4\ge0\)
canceled
\(x+7=0\Rightarrow x=-7\quad\text{(could be)}\) - \((x-9)\)
Numerator and denominator are simplified.
\(x-9\ne0\Rightarrow x\ne9\Rightarrow x,\ 9\ \text{(can not be)}\) - \(x^2+x+7\)
statements' Δ < 0: no root, always positive. Canceled.
- \((x+5)^{99}\)
Only the power of (x + 5)⁹⁹ statement will be canceled.
\(\samefrac{(x-2)}{x+5}\ge0\)
\((-\infty,-5)\cup[2,\infty)\)
The information above will be added to the solution set.
\(x=3\qquad x=-7\qquad x\ne9\)
\(\Rightarrow\{(-\infty,-5)\cup[2,\infty)\}\setminus\{9\}\)
1.
\(\samefrac{(x-2)(x+4)^2}{x^2}<0\)
\(\Rightarrow S.S.=?\)
2.
\(\samefrac{(x^2-4x+4)x^3}{3-x}\ge0\)
\(\Rightarrow S.S.=?\)
3.
\(\samefrac{x^2-6x+5}{(x-4)^2}<0\)
\(\Rightarrow S.S.=?\)