sir how to solve quadratic inequalities by wavy curve method - Maths - Complex Numbers and Quadratic Equations

Question

sir how to solve quadratic inequalities by wavy curve method

Shridhar Kaushik · Accepted Answer

Wavy curve method is used to find the solution set for a given inequality The steps involved in wavy curve method are : (1) Factorize the numerator and denominator into linear factors (2) Make coefficients of x positive in all linear factors (3) Equate each linear factor to zero and find the value of x in each case The values of x are called critical points (4) Mark these critical points on number line The "n" numbers of distinct critical points divide number line in (n + 1) sub-intervals (5) The sign of rational function in the right most interval is positive Alternate sign in adjoining intervals on the left (6) If a linear factor is repeated even times, then sign of function will not alternate about the critical point corresponding to linear factor in the question You can see the following example to get wavy curve method more clearly : Q Find the solution of the rational inequality given by : $\frac{3 x^{2} + 6 x - 15}{(2 x - 1) (x + 3)} \geq 1$ Solution: First of all, convert the given inequality to standard form f(x) â‰¥ 0 $\Rightarrow \frac{3 x^{2} + 6 x - 15}{(2 x - 1) (x + 3)} - 1 \geq 0 \Rightarrow \frac{x^{2} + x - 12}{(2 x - 1) (x + 3)} \geq 0 [taking L C M] \Rightarrow \frac{(x - 3) (x + 4)}{(2 x - 1) (x + 3)} \geq 0$ Critical points are â€“4, â€“3, $\frac{1}{2}$ and 3 The corresponding sign diagram is :

The solution of inequality is â€“ $x \in (- \infty,, - 4] \cup (- 3, \frac{1}{2}) \cup [3, \infty)$ We do not include "â€“3" and " $\frac{1}{2}$ " as they reduce the denominator to zero