Question

Help please solve %29%5E2%7D%20%5Cleq%200" id="TexFormula1" title="\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0" alt="\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$\displaystyle -\frac{1}{2} \leq x < 1$

Step-by-step explanation:

Inequalities

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

$\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0$

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

$\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0$

Simplifying by x-1 and taking x=1 out of the possible solutions:

$\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0$

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

$(6x^3+14x^2+10x+12)$

is always positive and doesn't affect the result. It can be neglected. The expression

$(x-\frac{1}{2})^2$

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

$\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0$

For the expression to be negative, both signs must be opposite, that is

$(x+\frac{1}{2})\geq 0, (x-1)$

Or

$(x+\frac{1}{2})\leq 0, (x-1)>0$

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

$\displaystyle -\frac{1}{2} \leq x < 1$

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

$\boxed{\displaystyle -\frac{1}{2} \leq x < 1 }$