Question

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

answers:
(−∞,−52]∪[23,∞)

(−∞,−23]∪[52,∞)

[−23,52]

[−32,14]

Answer 1

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

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)-(11)/(6)\cdot x -(10)/(6) \right)\geq 0`

Roots are found by Quadratic Formula:

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

`r_(1) = (5)/(2)` and
`r_(2) = -(2)/(3)`

Then, the factorized form of the inequation is:

`6\cdot \left(x-(5)/(2)\right)\cdot \left(x+(2)/(3) \right)\geq 0`

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

a)
`x-(5)/(2) \geq 0 \,\wedge\,x+(2)/(3)\geq 0`

`x \geq (5)/(2)\,\wedge\,x \geq-(2)/(3)`

`x \geq (5)/(2)`

b)
`x-(5)/(2) \leq 0 \,\wedge\,x+(2)/(3)\leq 0`

`x \leq (5)/(2)\,\wedge\,x\leq-(2)/(3)`

`x\leq -(2)/(3)`

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