Question

Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x

Answer 1

Answer:

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$.

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

$6\cdot x^{2} \geq 10 + 11\cdot x$

$6\cdot x^{2}-11\cdot x -10 \geq 0$

$6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0$

Roots are found by Quadratic Formula:

$r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}$

$r_{1} = \frac{5}{2}$ and $r_{2} = -\frac{2}{3}$

Then, the factorized form of the inequation is:

$6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0$

By Real Algebra, there are two condition that fulfill the inequation:

a) $x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0$

$x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}$

$x \geq \frac{5}{2}$

b) $x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0$

$x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}$

$x\leq -\frac{2}{3}$

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$.