Answer:
The solution of the inequation [tex]6\cdot x^{2} \geq 10 + 11\cdot x[/tex] is [tex]\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)[/tex].
Step-by-step explanation:
First of all, let simplify and factorize the resulting polynomial:
[tex]6\cdot x^{2} \geq 10 + 11\cdot x[/tex]
[tex]6\cdot x^{2}-11\cdot x -10 \geq 0[/tex]
[tex]6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0[/tex]
Roots are found by Quadratic Formula:
[tex]r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}[/tex]
[tex]r_{1} = \frac{5}{2}[/tex] and [tex]r_{2} = -\frac{2}{3}[/tex]
Then, the factorized form of the inequation is:
[tex]6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0[/tex]
By Real Algebra, there are two condition that fulfill the inequation:
a) [tex]x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0[/tex]
[tex]x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}[/tex]
[tex]x \geq \frac{5}{2}[/tex]
b) [tex]x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0[/tex]
[tex]x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}[/tex]
[tex]x\leq -\frac{2}{3}[/tex]
The solution of the inequation [tex]6\cdot x^{2} \geq 10 + 11\cdot x[/tex] is [tex]\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)[/tex].
