What is the soluton to the quadratic inequality?
\(6x^2 \geq 5-13x\)
\(\begin{array}{|rcll|} \hline \mathbf{6x^2} &\geq& \mathbf{5-13x} \quad | \quad -5+13x \\ 6x^2 +13x -5 &\geq& 0 \\ \overset{\text{quadratic formula}}{\ldots} \\ 6\left(x-\dfrac{1}{3}\right) \left(x+\dfrac{5}{2} \right) &\geq& 0 \quad | \quad : 6 \\\\ \mathbf{\left(x-\dfrac{1}{3}\right) \left(x+\dfrac{5}{2} \right)} &\geq&\mathbf{ 0 } \\ \hline \end{array} \)
We create a sign table:
\(\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Interval or position} : & \left(-\infty,-\dfrac{5}{2}\right) & -\dfrac{5}{2} & \left(-\dfrac{5}{2},\dfrac{1}{3}\right) & \dfrac{1}{3} & \left(\dfrac{1}{3},\infty\right) \\ \hline \text{sign of } \left(x+\dfrac{5}{2} \right): & - & 0 & + & + & + \\ \hline \text{sign of } \left(x-\dfrac{1}{3}\right): & - & - & - & 0 & + \\ \hline \text{sign of }\left(x-\dfrac{1}{3}\right) \left(x+\dfrac{5}{2} \right): & \color{red}+ & \color{red}0 & - & \color{red}0 & \color{red}+ \\ \hline \end{array} \)
We can read the result:
in the interval \(\left(-\infty,-\dfrac{5}{2}\right]\) and \(\left[\dfrac{1}{3},\infty\right)\) the left side of the inequality is positive or zero, and thus the inequality is true there.
The solution is: \(\mathbf{ x\leq -\dfrac{5}{2}}\) and \(\mathbf{x \geq \dfrac{1}{3}}\).
