Find all solutions to the inequality \(\frac{(2x-7)(x-3)}{x} \ge 0\) in interval notation
+26367 Find all solutions to the inequality \(\large{\dfrac{(2x-7)(x-3)}{x} \ge 0}\) in interval notation
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{(2x-7)(x-3)}{x}} &\ge& \mathbf{0} \quad | \quad \boxed{x\neq 0 !} \\\\ \dfrac{(2x-7)(x-3)}{x} &\ge& 0 \quad | \quad \times x^2 \\ \dfrac{(2x-7)(x-3)x^2}{x} &\ge& 0\times x^2 \\ (2x-7)(x-3)x &\ge& 0 \\ 2\left(x-\dfrac{7}{2}\right)(x-3)x &\ge& 0 \quad | \quad : 2 \\ \left(x-\dfrac{7}{2}\right)(x-3)x &\ge& 0 \\ \mathbf{ \left(x-\dfrac{7}{2}\right)(x-3)(x-0) } &\ge& \mathbf{0} \\ \hline \end{array}\)
We create a sign table:
\(\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Interval or position} : & (-\infty,0) & 0 & (0,3) & 3 & \left(3,\dfrac{7}{2}\right) &\dfrac{7}{2} & \left(\dfrac{7}{2},\infty\right) \\ \hline \text{sign of } (x-0): & - & 0 & + & + & + & + & + \\ \hline \text{sign of } (x-3): & - & - & - & 0 & + & + & + \\ \hline \text{sign of } \left(x-\dfrac{7}{2}\right): & - & - & - & - & - & 0 & + \\ \hline \text{sign of }\left(x-\dfrac{7}{2}\right)(x-3)(x-0): & - & \color{red}0 & \color{red}+ & \color{red}0 & - & \color{red}0 & \color{red}+ \\ \hline \end{array}\)
We can read the result:
in the interval \([0,3] \)and \(\Big[\dfrac{7}{2},\infty\Big)\) the left side of the inequality is positive or zero, and thus the inequality is true there.
But \(\boxed{x\neq 0 !}\) and the solution is: \(\Big(0,3\Big]\) and \(\Big[\dfrac{7}{2},\infty\Big)\)
