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Problem : Find solution of :
$$\frac{2}{1+x}+\frac{3}{1-x}<1$$
Solution : Rearranging, we have :
$$\Rightarrow \frac{2-2x+3+3x}{\left(1+x\right)\left(1-x\right)}-1<0$$ $$\Rightarrow \frac{5+x-\left(1-{x}^{2}\right)}{\left(1+x\right)\left(1-x\right)}<0$$ $$\Rightarrow \frac{{x}^{2}+x+4}{\left(1+x\right)\left(1-x\right)}<0$$
Now, polynomial in the numerator i.e. ${x}^{2}+x+4$ is positive for all real x as D<0 and a>0. Thus, dividing either side of the inequality by this polynomial does not change inequality. Now, we need to change the sign of x in one of the linear factors of the denominator positive in accordance with sign rule. This is required to be done in the factor (1-x). For this, we multiply each side of inequality by -1. This change in sign accompanies change in inequality as well :
$$\Rightarrow \frac{1}{\left(1+x\right)\left(1-x\right)}>0$$
Critical points are -1 and 1. Hence, solution of the inequality in x is :
$$x\in \left(-\infty ,-1\right)\cup \left(\mathrm{1,}\infty \right)$$
We have already discussed rational polynomial with repeated factors. We need to count repeated factors which appear in both numerator and denominator. If the linear factors are repeated even times, then we do not need to change sign about critical point corresponding to repeated linear factor.
Note : While working with rational function having repeated factors, we need to factorize higher order polynomial like cubic polynomial. In such situation, we can employ a short cut. We guess one real root of the cubic polynomial. We may check corresponding equation with values such as 1,2, -1 or -2 etc and see whether cubic expression becomes zero or not for that value. If one of the roots is known, then cubic expression is f(x) = (x-a) g(x), where "a" is the guessed root and g(x) is a quadratic expression. We can then find other two roots anlayzing quadratic expression. For example, ${x}^{3}-6{x}^{2}+11x-6=\left(x-1\right)({x}^{2}-5x+6)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$
Problem : Find interval of x satisfying the inequality given by :
$$\frac{\left(2x+1\right)\left(x-1\right)}{\left({x}^{3}-3{x}^{2}+2x\right)}\ge 0$$
Solution : We factorize each of the polynomials in numerator and denominator :
$$\Rightarrow \frac{\left(2x+1\right)\left(x-1\right)}{\left({x}^{3}-3{x}^{2}+2x\right)}=\frac{\left(2x+1\right)\left(x-1\right)}{x\left(x-1\right)\left(x-2\right)}$$
It is important that we do not cancel common factors or terms. Here, critical points are -1/2,1,0,1 and 2. The critical point "1" is repeated even times. Hence, we do not change sign about "1" while drawing sign scheme.
While writing interval, we drop equality sign for critical points, which corresponds to denominator.
$$-1/2\le x<0\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}2<x<\infty $$
$$[-1/2,0)\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}(2,\infty )$$
We do not include "-1" and "1" as they reduce denominator to zero.
We can treat polynomial inequality as rational inequality, because a polynomial function is a rational function with denominator as 1. Logically, sign method used for rational function should also hold for polynomial function. Let us consider a simple polynomial inequality, $f\left(x\right)=2{x}^{2}+x-1<0$ . Here, function is product of two linear factors (2x-3)(x+2). Clearly, x=3/2 and x=-2 are the critical points. The sign scheme of the function is shown in the figure :
Solution of x satisfying inequality is :
$$x\in \left(-2,\frac{3}{2}\right)$$
It is evident that this method is easier and mechanical in approach.
The term radical is name given to square root sign (√). A radical number is ${n}^{th}$ root of a real number. If y is ${n}^{th}$ root of x, then :
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